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1 

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6 

J 


THEORY   OF   STRUCTURES 


AND 


STRENGTH  OF  MATERIALS. 


WITH 


DIAGRAMS,   ILLUSTRATIONS,   AND  EXAMPLES. 


BY 

HENRY  T.  BOVEY,  M.A.,  D.C.L.,  F.R.S.C, 

PROFESSOR   OP  CIVIL  ENGINEEKING  AND   APPLIED  MECHANICS,    m'giI.L   UNIVBKSITV,    MONTREAL. 
MBMBBR  OP  THE  INSTITUTION   OP   CIVIL  BNGINBBRS  ;     MEMHBK   OF  THB   INSTITU- 
TION OF   MECHANICAL    ENGINEERS;    LATE   FELLOW  OF 
queens'  COLLEGE,   CAMBRIDGE  (ENG.). 


.f   -. 


NEW  YORK: 

JOHN    WILEY    &    SONS, 

68  East  Tenth  Street. 

1893. 


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Copyright,  1893, 

BV 

HENRY  T.  BOVEY. 


Febrir  Bbo*., 

Prinitr*, 

tX,  Pearl  Street, 

New  York. 


ROBSBT  DBCWIOHD, 

jClwtrotvper, 

411  &  i4S  Pearl  Btieetf 

Mew  York. 


TyEDICATED 

TO 

WClHsm  €.  mc-BonaW, 

WHOSE  BENEFACVION^   TO   M'gILL   UNIVERSITY 

HAVE  DONE   SO   MUCH   :o  ADVANCE   THE   CAUSE  OF 

SCIENTIFIC  EDUCATION. 


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PREFACE. 


The  present  work  treats  of  that  portion  of  Applied 
Mechanics  which  has  to  do  with  the  Design  of  Structures. 

Free  reference  has  been  made  to  the  works  of  other 
authors,  yet  a  considerable  amount  of  new  matter  has  been 
introduced,  as,  for  example,  the  Articles  on  "  Surface  Loading" 
by  Carus-Wilson,  "  The  Flexure  of  Columns  "  by  Findlay,  and 
"  The  Efficiency  of  Riveted  Joints  "  by  Nicolson ;  also  my 
own  Articles  on  "Maximum  Shearing  Forces  and  Bending 
Moments,"  "  The  Flexure  of  Long  Columns,"  "  The  Theorem 
of  Three  Moments,"  etc. 

I  am  much  indebted  to  Messrs.  C.  F.  Findlay  and  W.  B. 
Dawson  for  valuable  information  respecting  the  treatment  of 
Cantilever  Bridges,  Arched  Ribs,  and  the  Live  Loads  on  Bridges. 

To  Messrs.  J.  M.  Wilson,  P.  A.  Peterson,  C.  Macdonald, 
and  others,  many  thanks  are  due  for  data  respecting  the  Dead 
Weights  of  Bridges. 

I  am  under  deep  obligation  to  my  friend  Prof.  Chandler, 
who  has  kindly  revised  the  proof-sheets,  and  who  has  made 
many  important  suggestions. 

I  have  endeavored  so  to  arrange  the  matter  that  the 
student  may  omit  the  advanced  portions  and  obtain  a  com- 
plete elementary  course  in  natural  sequence. 

At  the   end    of   each    chapter,    a   number  of   Examples, 


Vi 


PREFACE. 


selected  for  the  most  part  from  my  own  experience,  are 
arranged  with  a  view  to  illustrating  the  subject-matter — an 
important  feature,  as  it  is  admitted  that  the  student  who  care- 
fully works  out  examples  obtains  a  mastery  of  the  subject 
which  is  otherwise  impossible. 

The  various  Tables  in  the  volume  have  been  prepared  from 
the  most  recent  and  reliable  results. 

A  few  years  ago  I  published  a  work  on  "  Applied  Me- 
chanics," consisting  mainly  of  a  collection  of  notes  intended 
for  the  use  of  my  own  students.  The  present  volume  may  be 
considered  as  a  second  edition  of  that  work,  but  the  subject- 
matter  has  been  so  much  added  to  and  rearranged  as  to  make 
it  almost  a  new  book.  I  venture  to  hope  that  this  volume 
may  prove  acceptable  not  only  to  students,  but  to  the  profes- 
sion at  large. 

Henry  T.  Bovey. 

McGiLL  College,  Montreal, 
November,  1892. 


CONTENTS. 


CHAPTER  I. 


Frameu  Structures. 

PACK 

Definitions i 

Frames  of  Two  or  More  Members 2 

Funicular  Polygon 3 

Polygon  of  Forces 4 

Line  of  Loads 5 

Mansard  Roof 6 

Non-closing  Polygons 7 

Funicular  Curve 10 

Centre  of  Gravity 11 

Moment  of  Inertia 12 

Cranes,  Jib 13 

"        Derrick 16 

"       Composite 31 

Shear  Legs 17 

Bridge  Trusses > 17 

Roof  Trusses iq 

King-po&t  Truss 21 

Incomplete  Frames 27 

Queen-post  Truss.,. 31 

Composite  Frames 32 

Roof-weights 37 

Wind-pressure 38 

Distribution  of  Loads 39 

Examples  of  Roof  Trusses 41 

Examples  of  Bridge  Trusses  (Fink,  Bollman,  Howe,    Bowstring,   Single- 
intersection,  etc) 52 

Method  of  Sections 62 

Piers 65 

vii 


':S1 


VIM 


CONTENTS. 


PACK 

Tables  of  Roof -weights  and  Wind-pressures 67 

Examples 70 

CHAPTER  II. 

Shearing  Forces  ano  Bending  Moments. 

Equilibrium  of  Beams 93 

Shearing  Force 95 

Bending  Moment 96 

Examples  of  Shearing  Force  and  Bending  Moment 97 

Relation  between  Shearing  Force  and  Bending  Moment 108 

Eftt:;t  of  Live  (or  Rolling)  Load iil 

Graphical  Representation  of  Moment  of  Forces  with  Respect  to  a  Point.. .  116 

Relation  between  Bending  Moments  and  Funicular  Polygon 118 

Maximum  Shear  and  Maximum  Bending  Moment  at  any  Point  of  an  Arbi- 
trarily Loaded  Girder 121 

Hinged  Girders 136 

Examples 131 

CHAPTER  IIL 

General  Pkincii'LEs,  etc. 

Definitions 140 

Stress,  Simple 140 

"      Compound 140 

Hooke's  Law 141 

Coefficient  of  Elasticity 141 

Poisson's  Ratio 143 

Effect  of  Temperature 142 

Specific  Weight. 143 

Limit  of  Elasticity 143 

Breaking  Stress 147 

Dead  and  Live  Loads 143 

Repeated  Stress  Effect 145 

WOhler's  Lxperimenis 145 

Testing  of  Metals 147 

Launhardt's  Formula  159 

Wey ranch's  Formula 1 53 

Unwin's  Formula 159 

Flow  of  Solids 162 

Work,  Internal  and  External 168 

Energy,  Kinetic  and  Potential 167 

Oblique  Resistance 169 

Values  of  k 174 

Momentum.       Impulse 176 

Angular  Momentum 177 


CONTENTS. 


IX 


PACK 

Useful  Work.     Waste  Work 178 

Centrifugal  Force 181 

Impact 184 

Extension  of  a  Prismatic  Bar 189 

Oscillatory  Motion  of  a  Weight  at  the  End  of  a  Vertical  Elastic  Rod 190 

Inertia  198 

Balancing 198 

Curves  of  Piston  Velocity 205 

Linear  Diagrams  of  Velocity 206 

Curves  of  Crank-eflfort. . .    207 

Curves  of  Energy 207 

Fluctuation  of  Energy '. . .   207 

Tables  of  Strengths,  Elasticities,  and  Weights  of  Materials 210 

Tables  of  the  Breaking  Weights  and  Coefficients  of  Bending  Strengths  of 

Beams .  713 

Table  of  the  Weights  and  Crushing  Weights  of  Rocks,  etc 4 

Table  of  Expansions  of  Solids .!I3 

Examples 2t6 


CHAPTER  IV. 

Stresses,  Strains,  Earthwork,  and  RtfAiNiNO  \Valls. 

Internal  St:      as 2'^e 

Simple  Strain 235 

Compound  Strain 236 

Principal  Stresses 240 

Curves  of  Maximum  Shear  and  Normal  Intensity 240 

Combined  Bending  and  Twisting  Stresses 244 

Combined  Longitudinal  and  Twisting  Stresses 247 

Conjugate  Stresses 247 

Relation  between  Principal  and  Conjugate  Stresses 247 

Ratio  of  Conjugate  Stresses 250 

Relation  between  Stress  and  Strain 231 

Rankine's  Earthwork  Theory 255 

Pressure  against  a  Veriiv:al  Plane 257 

Earth  Foundations 258 

Retaining  Walls 260 

Retaininij  Walls.     Conditions  of  Equilibrium 260 

Rankine  3  Earthwork  Theory  applied  to  Retaining  Walls 264 

Line  of  Rupctire  265 

Practical  Rules  respecting  Retaining  Walls 267 

Reservoir  Walls 271 

General  Case  of  Reservoir  Walls 275 

General  Equations  of  Stress 276 

Ellipsoid  of  Stress a8i 

Stress-strain  Equations 281 


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CONTENTS. 


1 1 


Isotropic  Bodies  283 

Relation  between  A,  \,  and  G 285 

Traction 287 

To.sion 268 

Woric  done  in  the  Small  Strain  of  a  Body  (Clapeyron) 292 

Examples 294. 


CHAPTER  V.                                  .    i     ,  . 

Friction. 

Friction 300 

Laws  of  Friction 300 

Inclined  Plane 301 

We 


/edge 


302 


Screws 306 

Endless  Screw 309 

Rolling  Friction 310 

Journal  Friction 312 

Pivot 316 

Cylindrical  Pivot 316 

Wear 318 

Conical  Pivot 319 

Schiele's  Pivot  {anti-friction) 320 

Belts  and  Ropes 321 

Brakes 323 

Effective  Tension  of  a  Belt 324 

Effect  of  High  Speed 325 

Slip  of  Belts 326 

Prony's   Dynamometer 327 

StifTness  of  Belts  and  Ropes 327 

Wheel  and  Axle 329 

Toothed  Gearing 331 

Be vel- wheels 335 

Efficiency  of  Mechanisms. . .    335 

Table  of  Coefficients  of  Journal  Friction 336 

Examples 337 


CHAPTER  VI. 

Transverse  Strength  of  Beams. 

Elastic  Moment 340 

Moment  of  Resistance 340 

Neutral  Axis 340 

Transverse  Deformation 344 

Coefficient  of  Bending  Strength 344 


CONTENTS.    .  » 

FAGE 

Equalization  of  Stress 349 

Surface  Loading 350 

Effect  of  Bending  Moment  in  a  Plane  which  is  not  a  Principal  Plane 354 

Springs 355 

Beams  of  Uniform  Strength 358 

Flanged  Girders 365 

Classification  of  Flanged  Girders 365 

Equilibrium  of  Flanged  Girders 366 

Moments  of  Inertia  of  I  and  other  Sections 371 

Design  of  a  Girder  of  I-section 381 

Deflection  of  Girders 384 

Camber , 3S7 

Stiffness 389 

Distribution  of  Shearing  Stress 391 

Beam  acted  upon  by  Forces  Oblicue  to  its  Direction 396 

Similar  Girders 401 

Allowance  to  be  made  for  Weight  of  Bcim 405 

Examples 407 


CHAPTER  VII. 

Transverse  SxRENfiTH  ok  Beams— {Co«//w«<f</.) 

General  Equations  428 

Interpretation  of  the  General  Equations  432 

Examples — Cantilever 435 

"  Girder  upon  Two  Supports 439 

"  "      fixed  at   One  End   axid  resting   upon    Support    at  the 

Other 442 

••  "     fixed  at  Both  Ends 445 

••  "     upon  Two  Supports  not  in  the  same  Horizontal  Plane.  446 

"  Neutral  Axis  of  Arbitrarily  Loaded  Girders 448 

"  Cantilever  with  Varying  Section 455 

"  Girder  Encastr^  at  the  Ends 458 

Springs 456 

Work  done  in  bending  a  Beam 460 

Transverse  Vibrations  of  a  Beam  supported  at  Both  Ends 461 

Imperfect  Fixture 461 

Continuous  Girders 4C3 

Theorem  of  Three  Moments 463 

Swing-bridges 470 

Maximum  Bending  Moment  at  the  Points  of  Support  of  a  Continuous 

Girder  of  n  Spans 475 

General  Theorem  of  Three  Moments 484 

Comparative  Merits  of  Continuous  Gir  !ers 486 

Examples 490 


Xlt 


CONTENTS. 


•'••■■  CHAPTER  VIII.  ■  ^ 

.'■■-.,•■,  •-■...,-.•''-. 

Pillars. 

PACB 

Classification  of  Pillars 513 

Form  "        "      514 

Failure  "        "      515 

Uniform  Su ^ss S'^ 

Uniformly  Varying  Stress  517 

Hodgkinson's  Formulae 520 

Gordon's  Formula 523 

Values  of  the  CoeflScients  (a  and/)  in  Gordon's  Formula 523 

Graphical  Representation  of  Strength  of  Pillars 524 

Rankine's  Modification  of  Gordon's  Formula 526 

Formula  for  Safe  Working  Stress 536 

Value  of  "  Radius  of  Gyration  "  for  Dififerent  Sections 536 

American  Iron  Columns 533 

Long  Thin  Pillars 534 

Long  Columns  of  Uniform  Section  (Euler's  Theory) 538 

Resistance  of  Columns  to  Buckling  (Weyrauch's  Theory) 550 

Baker's  Formulae.... 549 

Flexure  of  Columns 554,  557 

Examples 563 


CHAPTER  IX. 

Torsion. 

Definition 568 

Coulomb's  Laws 568 

Totsional  Strength  of  Shafting 569 

St.  Venant's  Results 573 

Torsional  Rupture 573 

Resilience  of  Shafting 574 

Effect  of  Combined  Bending  and  Twisting 574 

Distance  between  Bearings  for  Shaftinc; 575 

Efficiency  of  Shafting 577 

Spiral  Springs 577 

Figures  illustrating   the  Distortion  produced  by  twisting  Round,  Square, 

and  Rectangular  Iron  Bars 585(1,  585^ 

Examples 580 


CHAPTER  X. 

Cylindrical  and  Spherical  Boilers. 

Cylinders 586 

Efficiency  of  Riveted  Joints  in  Boilers 587 


CONTENTS. 


xin 


Thick  Hollow  Cylinder 588 

Spherical  Shells 591 

Practical  Formulae 592 

Examples. . , 594 


CHAPTER  XI. 
Bridges. 

Classification , 597 

Curved  and  Horizontal  Flanges 597 

Depth  of  Girders  (or  Trusses) 597 

Position  of  Platform 598 

Comparative  Advantages  of  Two,  Three,  and  Four  Main  Girders 600 

Dead  Load 600 

Live  Load 600 

Trellis  (or  Lattice)  Girder 600 

Warren  Girder 603 

Howe  Truss 611 

Single-intersection  Truss 616 

Double-intersection  Truss 616 

Whipple  Truss 618 

Linville  Truss 618 

Post  Truss 618 

Quadrangular  Truss 618 

Bowstring  Truss 618 

Bowstring  Truss  with  Isosceles  Bracing  624 

Bowstring  Suspension-bridge   (Lenticular  Truss) 626 

Cantilever  Trusses 627 

Curve  of  Cantilever  Boom 634 

Deflection  of  Cantilevers 638 

Rollers 639 

Wind-pressure 651 

Regulations  respecting  Wind-pressure 653 

Lateral  Bracing 654 

Chords 655 

Stringers 656 

Maximum  Allowable  Working  Stresses 657 

Camber 659 

Rivet  Connections  between  Flange  and  Web \. 660 

Eye-bars,  Pins,  and   Rivets 661 

Steel  Eyebai's 665 

Rivets 666 

Dimensions  of  Rivets 667 

Strength  of  Punched  and  Drilled  Plates 668 

Riveted  Joints 668 

Theory  of  Riveted  Joints 6,» 


¥P 


XIV 


CONTENTS. 


»  PAGB 

Covers 675 

Efficiency  of  Riveted  Joints 676 

Tables  of  Weights  of  Actual  Bridges 682-687 

Table  of  Loads  for  Highway  Bridges 633 

Examples 689-702 


CHAPTER  XII. 

Suspension-bridges.  i     ,,      , 

Cables 703 

Anchorage '. 704 

Suspenders 706 

Curve  of  Cable  {catenary) 706 

Link  "Cable" 709 

Length  of  Cable 712 

Weight  of  Cable 713 

Deflection  due  to  Change  of  Length 714 

Pressure  upon  Piers 718 

Stiffening  Truss 719 

Stiffening  Truss  hinged  at  the  Centre 725 

Suspension-bridge  Loads 730 

Modifications  of  the  Suspension-bridge  proper 731 

Examples 734-739 


CHAPTER  XIII. 

Arched  Ribs. 

Definitions 740 

Equilibrated  Polygon  (Line  of  Resistance) 741 

Polygon  of  Pressures 743 

Linear  Arch 743 

Conditions  of  Equilibrium 745 

Joint  of  Rupture 747 

Minimum  Thickness  of  Abutment 749 

Empirical  Formulae 750 

Linear  Arch  in  Form  of  a  Parabola 750 

"     "       "       "     Transformed  Catenary 750 

" '     Circular  Arc 753 

"     "        "       "     Elliptic  Arc 753 

Hydrostatic  Arch 757 

Geostaiic  Arch 759 

General  Arch  Theory 760 

Arched   Ribs  762 

Bending  Moment  and  Thrust  at  any  Point  of  an  Arched  Rib 763 

Rib  with  Hinged  Ends.. .    . .  764 


CONTENTS. 


XV 


PAGE 

Semicircular  Rib  with  Hinged  Ends 765 

Graphical  Determination  of  the  Thrust  at  any  Point  of  an  Arched  Rib 767 

Rib  in  Form  of  a  Circular  Arc 769 

Rib  with  Fixed  Ends 771 

'         "    in  Form  of  a  Circular  Arc 773 

"      "        "         "     "       "        "     Semicircle 775 

"      "        "         "     "       "        "     Parabola 775 

Effect  of  a  Change  of  Temperature 777,  786 

Deflection  of  an  Arched  Rib 780,  802 

Elementary  Deformation  of  an  Arched  Rib 781 

Rib  of  Uniform  Stiffness 788 

Parabolic  Rib  of  Uniform  Depth  and  Stiffness 789,  795,  8cx) 

Arched  Rib  of  Uniform  Stiffness  with  Fixed  Ends 804 

Stresses  in  Spandril  Posts  and  Diagonals 804 

Maxwell's  Method  of  Determining  the  Stresses  in  a  Framed  Arch 806 

Examples 809-812 


ITS 


II 


I! 

» I 


THEORY    OF   STRUCTURES. 


CHAPTER   I. 


FRAMES   LOADED   AT   THE  JOINTS. 


I.  Definitions. — Frames  are  rigid  structures  composed  of 
straight  struts  and  ties,  jointed  together  by  means  of  bolts, 
straps,  mortises  and  tenons,  etc.  Struts  are  members  in  com- 
pression, ties  members  in  tension,  and  the  term  brace  is  applied 
to  either. 

The  external  forces  upon  a  frame  are  the  loads  and  the 
reactions  at  the  points  of  support,  from  which  may  be  found 
the  resultant  forces  at  the  joints.  The  greatest  care  should  be 
exercised' in  the  design  of  the  joints.  The  resultant  forces 
should  severally  coincide  in  direction  with  the  axes  of  the 
members  upon  which  they  act,  and  should  intersect  the  joints 
in  their  centres  of  gravity.  Owing  to  a  want  of  homogeneity 
in  the  material,  errors  of  workmanship,  etc.,  this  coincidence  is 
not  always  practicable,  but  it  should  be  remembered  that  the 
smallest  deviation  introduces  a  bending  action.  Such  an 
action  will  also  be  caused  by  joint  friction  when  the  frame  is 
insufficiently  braced.  The  points  in  which  the  lines  of  action 
of  the  resultants  intersect  the  joints  are  also  called  the  centres 
of  resistance,  and  the  figure  formed  by  joining  the  centres  of 
resistance  in  order  is  usually  a  polygon,  which  is  designated  the 
line  of  resiiianec  oi  the  frame. 

The  position  of  the  centres  should  on  no  account  be  allowed 
to  vary.  Ii  is  assumed,  and  is  practically  true,  that  the  joints 
of  a  frame  are  flexible,  and  that  the  frame  under  a  given  load 


2  '  THEORY  OF  STRUCrV RES. 

does  not  sensibly  change  in  form.  Thus  an  individual  mem- 
ber is  merely  stretched  or  compressed  in  the  direction  of  its 
length,  i.e.,  along  its  line  of  resistance,  while  the  frame  as  a 
whole  may  be  subjected  to  a  bending  action. 

The  term  truss  is  often  applied  to  a  frame  supporting  a 
weight, 

2.  frame  of  Two  Members.  —  OA,  OB  are  two  bars 
jointed  at  0  and  supported  at  the  ends  A^  B.     The  frame  in 


Fig.  I. 


Fig.  3. 


Fig.  3. 


Fig.  I  consists  of  two  ties,  in  Fig.  3  of  two  struts,  and  in  Fig. 
2  of  a  strut  and  a  tie. 

Let  P  be  the  resultant  force  at  the  joint,  and  let  it  act  in 
the  direction  OC.  Take  OC  equal  to  P  in  magnitude,  and 
draw  CD  parallel  to  OB.  OD  is  the  stress  along  OA,  and  CD 
is  that  along  OB. 

Let  tht  angle  AOB  —  a,  and  the  angle  COD  =  /?. 

Let  6'i ,  5,  be  the  stresses  along  OA,  OB,  respectively. 


5.  _  O^  _  sin(Q-  -  /?) 
P~OC~      sin  a      ' 


and    ^^  =  ^  = 


CD 

OC 


sin  /3 
sin  a 


3.  Frame  of  Three  or  More  Members —Let  A,A,A, . . . 
be  a  polygonal  frame  jointed  at  A^,  A^,  A^,  .  .  .  Let/',, 
/*,,  P,,  .  .  .  be  the  resultant  forces  at  the  joints  A^,  A,,  A^, 
.  .  .  ,  respectively.  Let  5,,  S^,  5,,  .  .  .  be  the  forces  along 
A^A,,  A,A^,  .  .  .  ,  respectively.  , 

Consider  the  joint  A^. 

The  lines  of  action  of  three  forces,  P, ,  5, ,  and  5,,  intersect 
in  this  joint,  and  the  forces,  being  in  equilibrium,  may  be 
represented  in  direction  and  magnitude  by  the  sides  of   the 


FRAMES  LOADED  AT   THE  JOINTS,  3 

triangle  Os^s^,  in  which  ^,5,  is  parallel  to  /*,,  Os^  to  5,,  and  Os^ 

to  5,.  .      .  i:' 

Similarly, /!,,  5,,  5,  maybe  represented  by  the  sides  of 
the  triangle  OSyS,  which  has  one  side,  Os^,  common  to  the 
triangle  Os^s^,  and  so  on. 

Thus  every  joint  furnishes  a  triangle  having  a  side  common 
to  each  of  the  two  adjacent  triangles,  and  all  the  triangles  to- 
gether form  a  closed  polygon  s^s^s^.  .  .  The  sides  of  this 
polygon  represent  in  magnitude  and  direction  the  resultant 


Fig.  4. 


forces  at  the  joints,  and  the  radii  from  the  pole  O  to  the  angles 
s^s^s^ ,  .  .  .  represent  in  magnitude,  direction  and  character, 
the  forces  along  the  several  sides  of  the  frame  A^A^A^.  .  , 
The  polygon  A^A^A^  .  .  .is  the' line  of  resistance  of  the 
frame,  and  is  called  the  funicular  polygon  of  the  forces  P, ,  P, , 
J\,  .  .  .  with  respect  to  the  pole  O. 

The  two  polygons  are  said  to  be  reciprocal,  and,  in  general, 
two  figures  in  graphical  statics  are  said  to  be  reciprocal  when 
the  sides  in  the  one  figure  are  parallel  or  perpendicular  to  cor- 
responding sides  in  the  other.  -   ,.  '  •' 

A  triangle  or  polygon  is  also  said  to  be  the  reciprocal  of  a 
point  when  its  sides  are  parallel  or  perpendicular  to  correspond- 
ing lines  radiating  from   the  point.     Thus  the  triangle  Os^s^  is 


Mil 


THEORY  GF  STRUCTURES. 


the  reciprocal  of  the  point  A^,  and  the  polygon  A^A^,  .  .  . 
is  the  reciprocal  of  the  point  0. 

If  more  than  fttfo  members  meet  at  a  joint,  or  if  the  joint  is 
subjected  to  more  than  o/ie  load,  the  resulting  force  diagram 
will  be  a  quadrilateral,  pentagon,  hexagon,  .  .  .  according  as 
the  number  of  members  is  3,  4,  5,  .  .  .  or  the  number  of  loads 

2,  3.  4.  •  •  . 

In  practice  it  is  usually  required  to  determine  the  stresses 
in  a  number  of  members  radiating  from  a  joint  in  a  framed 
structure.  If  the  reciprocal  of  the  joint  can  be  drawn,  its 
sides  will  represent  in  direction  and  magnitude  the  stresses  in 
the  corresponding  members. 

Corollary. — The  converse  of  the  preceding  is  evidently  true. 
For  if  a  system  of  forces  is  in  equilibrium,  the  polygon  of 
forces  s^s^^ .  .  .  must  close,  and  therefore  the  polygon  which 
has  its  sides  respectively  parallel  to  the  radii  from  a  pole  O  to 
the  angles  j, ,  J,,  j, ,  .  .  .  and  which  has  its  angles  upon  the 
lines  of  action  of  the  forces,  must  also  close. 

Example  i.  Let  (9  be  a  joint  in  a  framed  structure,  and 
let  Os^,  Os„  OSf,  ...  be  the  axes  of  the  members  radiating 
from  it.  The  polygon  A^A^A^  ...  is  the  reciprocal  of  O,  the 
side  AjA^  representing  the  stress  along  Os^,  the  side  A^A^  that 
along  Os.^,  etc. 

Ex.  2.  Let  the  resultant  forces  at  the  joints  be  paral- 
lel.    The  polygon  of   forces  becomes  the  straight  line  s^s^. 


m 


Fig.  6. 


FRAMES  LOADED  AT   THE  JOLVTS. 


which  is  often  termed  the  line  of  loads.  Thus,  tlie  forces  /*, . 
/'„...  Pt  are  represented  by  the  sides  j, J,,  s^s^,  .  .  .  s^s^,  which 
are  in  one  straight  Hne  closed  by  s^s^  and  s^s^,  representing  the 
remaining  forces  /*,  and  /*,,  and  tiie  triangles  Os^s.,,  Os,s^,  .  .  . 
are  the  reciprocals  of  the  points  A^,  A,.  .  .  .  Draw  OH  per- 
pendicular to  s,s^.  Tile  projection  of  each  of  the  lines  Os,, 
Os,,  Os^,  .  .  .  perpendicular  to  s^s^  is  the  same  and  equal  to 
OH,  which  therefore  represents  in  magnitude  and  direction 
the  stress  which  is  the  same  for  each  member  of  the  frame. 

Let    «',,«',,  <f,,  •  .  •  be   the    inclinations  of  the  members 
A^A,,  A,A^,  .  .  .  respectively,  to  the  line  of  loads.  Then 

OH  =  Hs^  tan  «,  =  Hs^  tan  or, ; 
.*.  OH{cot  a,  +  cot  a,)  =  Hs,  -\-  Hs^  =  J,.y, 

=  /',  +  /'.  +  /'.  +  /'.  =  /'.  +  /'., 

and  OH,  in  direction  and  magnitude,  is  equal  to  the  stress 
common  to  each  member.  Also,  the  stress  in  any  member, 
e.g.,  A^A^  =  Os^  =  OH  cosGC  a^ . 

Corollary. — Let  the  resultant  forces  at  the  joints  /!,,  A^ 
be  inclined  to  the  common  direction  of  the  remaining  forces, 
and  act  in  the  directions  shown  by  the  dotted  lines.  Let  /*/, 
/'/  be  the  magnitudes  of  the  new  forces;  draw  ^,5/  parallel  to 
the  direction  of  P^'  so  as  to  meet  Os^  in  s^  ;  join  s^s^'.  Since 
there  is  equilibrium,  s^s^  must  be  parallel  to  the  line 
of  action  of  P^  .     Thus,  sJs^s^  is  the  force  polygon. 

Ex.  3.  The  forces,  or  loads,  /!,,  P^,  .  .  .  P^  are 
generally  vertical,  while  /*, ,  /*,  are  the  vertical  re- 
actions of  the  two  supports. 

Suppose,  e.g.,  that  A^A^  ...  y^,  is  a  rope  or  chain 
suspended  from  the   points  ^,,  A^,  in  a  horizontal  *"■ 
plane  and  loaded  at  A^A^  .  .  .  with  weights  /',,/',,...  h- 
The  chain  will  hang  in  a  form  dependent  upon  the  * 
magnitude  of  these  weights.     The  points  H  and  5,    * 
will  coincide,  and  OH  will  represent  the  horizontal  *'' 
tension  of  the  chain. 

Let  the  polygon  A^A^  .../},  be  inverted,  and  let  the  rope 
be  replaced  by  rigid    bars,  A,A,,  A,A^.  .  .     The  diagram  of 


Fig.  8. 


V  ! 


THEORY  OF  STRUCTURES. 


-.1  ..ii 


forces  will  remain  the  same,  and  the  frame  will  be  in 
equilibrium  under  \.\\c given  loads.  The  equilibrium,  however, 
is  unstable  as  the  chain,  and  consequently  the  inverted  frame 
will  change  form  if  the  weights  vary.  Braces  must  then  be 
introduced  to  prevent  distortion. 


D/'?3_ 


Fig.  9. 


Fig.  10. 


Take  the  case  of  a  frame  DCBA  .  .  .  symmetrical  with 
respect  to  a  vertical  through  A,  and  let  the  weights  at  A,B,  C, 
.  .  .  be  W^,  U\,  IFj,  .  .  .  ,  respectively. 

Drawing  the  stress  diagram  in  the  usual  manner,  OH  rep- 
resents the  horizontal  thrust  of  tlie  frame. 

The  portions  s^s^,  s^s^,  ...  of  the  line  of  loads  give  a 
definite  relation  between  the  weights  for  which  the  truss  will 
be  stable.  The  result  may  be  expressed  analytically,  as 
follows : 

Let  a,,  a,,  a,,  ...  be  the  inclinations  of  AB,  BC,  CD,  .  .  .  , 
respectively,  to  the  horizontal. 

Let  the  horizontal  thrust  OH  =  H.  Then 

•  \ 

H=  ^  cot  a,  =  {^  ^+  W^,)cot  «,=(^+  W,-^  w}jco\.  a,^... 


il 

w 


•'  I IH! 


cot  tr,  =  3  cot  «',  =  5  cot  0-3  =  .  .  . 

If  there  are  two  bars  only,  viz.,  AB,  BC,  on  each  side  of  the 
vertical  centre  line,  the  frame  will  have  a  double  slope,  and  ia 
this  form  is  employed  to  support  a  Mansard  rooi.  ' 


FKAMES  LOADED   AT    THE  JOINTS.  7 

4.  Non-closing  Polygons. — Let  a  number  of  forces  /*, , 
/•,,/',,...  act  upon  a  structure,  and  let  these  forces,  taken  in 
order,  be  represented  in  direction  and  niatjnituile  by  the  sides 
of  the  unclosed  figure  MNPQ  .  .  .     This  figure  is  the  unclosed 


polygon  of  forces,  and  its  closing  line  T^^  represents  in  direction 
and  magnitude  the  resultant  of  the  forces  P^ ,  P^ ,  P^ ,  .  .  . 

For  PM  is  the  resultant  of  /*,  and  /*,,  and  may  replace 
them  ;  QM  may  replace  PM  and  P^,  i.e.,  /*,,  P^,  and  P^;  and 
so  on. 

Take  any  point  O  and  join  OM,  ON,  OP,  .  .  . 

Draw  a  line  AB  parallel  to  OJf  and  intersecting  the  line  of 
action  of  /*,  in  any  point  B.  Through  B  draw  73C  parallel  to 
ON  and  cutting  the  line  of  action  of  P^  in  C.  Similarly,  draw 
CD  parallel  to  OP,  DE  to  OQ,  EF  to  OR,  .  .  .  The  figure 
ABCD  ...  is  called  the  funiailar  polygon  of  the  given  forcer, 
with  respect  to  \.\\e.  pole  O.  The  position  of  the  pole  O  is  arbi- 
trary, and  therefore  an  infinite  number  of  funicular  polygons 
may  be  drawn  with  different  poles. 

Also  the  position  of  the  point  B  in  the  line  of  action  of  /' 
is  arbitrary,  and  hence  an  infinite  number  of  funicular  polygons 
with  their  corresponding  sides  parallel,  i.e.,  an  infinite  number 
of  similar  funicular  polygons,  may  be  drawn  with  the  same  pole. 


\\      >l 

*  J      i\ 


.    1 


«  THEORY  OF  STRUCTURES. 

5.  To  show  that  the  Intersection  of  the  First  and  Last 
Sides  of  the  Funicular  Polygon  (i.e.,  the  Point  G)  is  a  Point 
on  the  Actual  Resultant  of  the  System  of  Forces  P,,  P, , 

Pj,  .  .  . — First  consider  two   points  T^, ,  /*,,  MNP  being  the 
force  and  ABCD  the  funicular  polygon. 

Let  yJZ>,  DC,  the  first  and  last  sides  of  the  latter,  be  pro- 


FiG.  la. 


duced  to  meet  in  g^^  also  let  Z?C produced  meet  the  line  of 
action  of  /*,  in  H. 

Produce  t^/'and  MN  to  meet  in  K. 

Let  the  lines  of  action  of  /",  and  /*,  meet  in  L. 

By  similar  triangles, 


H 


P. 


Hence 


or 


KP 
KN 

HC 

~  HL' 

KN 
KO  ~ 

HB 
HC 

KO 

KM~ 

HB' 

KP  KN  KO 
KN  KO  KM  ~ 

HC  HB  Hg, 
'  HL  HC  HB  ' 

KP 
KM~ 

'  HV 

pnd  therefore,  since  the  angle  H  is  equal  to  the  angle  K,  the 
line  PM\s  parallel  to  the  line  Lg^. 


FRAMES  LOADED  A  T   THE  JOINTS. 


But  PM  represents  in  magnitude  the  resultant  of  the  forces 
P,,  P^,  and  is  parallel  to  it  in  direction. 

Therefore  Lg^  is  also  parallel  to  the  direction  of  the  re- 
sultant. 

But  L  is  evidently  a  point  on  the  actual  resultant  of  P^ ,  P, . 
Hence ^,  must  be  a  point  on  this  resultaiit. 

7\V.i7,  let  there  be  three  forces,  P^,  P^,  P^. 

Replace  P, ,  P^  by  their  resultant  X  acting  in  the  direction 
Lg^.  The  force  and  funicular  polygons  for  the  forces  A' and 
P,  are  evidently  MPQ  and  Ag^DE,  respectively  ;  and  g^ ,  the 
point  of  intersection  of  Ag^  and  ED  produced,  is,  as  already 
proved,  a  point  on  the  actual  resultant  of  X  and  P^ ,  i.e.,  of 
P, ,  P, ,  and  P3 . 

Hence  the  Jirst  and  /ast  sides,  AB,  ED,  of  the  funicular 
polygon  ABCDE  of  the  forces  P^,  P^,  P^ ,  with  respect  to  the 
pole  O,  intersect  in  a  point  which  is  on  the  actual  resultant  of 
the  given  forces. 

The  proof  may  be  similarly  extended  to  four,  five,  and  any 
number  of  forces. 

If  the  forces  are  all  parallel,  the  force  pul\-gon  of  the  two 
forces  P^ ,  Pj  becomes  a  straight  line,  MNQ.    Draw  the  funicular 


Fig.  13. 


polygon  ^5C/?  as  before,  and  through  ^if,,  the  intersection  of  the 
y?rj/and  /^/j/ sides,  dravv^'-,  Kparallel  tOiT/(2.  and  cutting  Z?C"  in  Y. 
By  similar  triangles, 


ON  ~  ON  ~  BY' 


and 


ON  ~  ON  ~  CY  ' 


P. 


CY^ 
BY' 


I! 


10 


THEORY  OF  STRUCTURES. 


'      ^1 


:, 


Hence  Vg^,  which  is  parallel  to  the  direction  of  the  forces /*,, 
/*,,  divides  the  distance  between  their  lines  of  action  into  seg- 
ments which  are  inversely  proportional  to  the  forces,  and  must 
therefore  be  the  line  of  action  of  their  resultant.  The  proof 
may  be  extended  to  any  number  of  forces,  as  in  the  preceding. 

Funicular  Curve. — Let  the  weights  upon  a  beam  AB  become 
infinite  in  number,  and  let  the  distances  between  the  weights 
diminish  indefinitely. 

The  load  then  becomes  continuous,  and  the  funicular  poly- 
gon is  a  curve,  called  the  funicular  curve. 

The  equation  to  this  curve  may  be  found  as  follows : 

Let  the  tangents  at  two  consecutive  points  /*and  Q  meet 
in  R.  This  point  is  on  the  vertical  through  the  centre  of 
gravity  of  the  load  upon  the  portion  MN  of  the  beam. 


A     X     Mt/JN 


B 


Fig.  14.  Fig.  15. 

Let  17/  be  the  hne  of  loads,  and  let  OS,  OT  ho.  the  radial 
lines  from  O,  the  pole,  parallel  to  the  tangents  at  P  and  Q. 
Take  ^  as  the  origin. 

Let  ^  be  the  inclination  of  the  tangent  at  P  to  the  beam, 
and  let  the  polar  distance  OV  =  p. 

wdx  =  the  load  upon  the  portion  JAV.     'riicn 

wc^x  =  ST=S1'-  TV  =p  tan  ^  -  /  tan  Kj^  +  d^) 

==  —  pdti,  approximately. 


de 


d'y 


''=^dv  =  Kix- 


since     ^  = 


dx' 


Integrating  twice, 

c^  and  i*,  being  constants  of  integration. 


CENTRES  OF  GRAVITY. 


II 


If  the  intensity,  w,  of  the  load  is  constant, 

py  ^  - -^ -\- c,x -^ c^, 

and  the  curve  is  a  parabola. 

6.  Centres  of  Gravity. — Let  it  be  required  to  determine 
the  centre  of  gravity  of  any  plane  area  symmetrical  with  re- 
spect to  an  axis-<YLY.  Divide  the  area  into  suitable  elementary 
areas  ^,,  ^,,  ^,,  .  .  .  having  known  centres  of  gravity. 


Fig. 


'7- 


Draw  the  force  (the  line  i«)  and  funicular  polygons  corre- 
sponding to  these  areas,  and  let  ghc  the  point  in  which  the  first 
and  last  sides  of  the  funicular  polygon  meet.  The  line  drawn 
through  g  parallel  to  \n  must  pass  through  the  centre  of  gravity 
of  all  the  elementary  areas  and.  therefore,  of  the  whole  area. 
Hence  it  is  the  point  (7  in  which  this  line  intersects  the  axis  XX. 

Rail  and  similar  sections  may  be  divided  into  elementary 
areas  by  drawing  a  number  of  parallel  lines  at  right  angles  to 
the  axis  of  symmetry,  and  at  such  distances 
apart  that  each  elementary  figure  may,  with- 
out sensible  error,  be  considered  a  rectangle 
of  an  area  equal  to  the  product  of  its  breadth 
by  its  mean  height. 

In  the  case  of  a  very  irregular  section,  an 
accurate  template  of  the  section  may  be  cut 
out  of  cardboard  or  thin  metal.  If  the  tem- 
plate is  then  suspended  from  a  pin  through  a  point  near  the 


J 


Fig.  i8. 


i  : 


12 


THEORY  OF  STRUCTURES. 


edge,  the  centre  of  gravity  of  the  section  wil'  He  in  the  vertical 
through  the  pin.  By  changing  the  point  of  suspension,  a  new 
line  in  which  the  centre  of  gravity  lies  may  be  found.  The 
intersection  of  the  two  lines  must,  therefore,  be  the  centre  of 
gravity  required.  Another  method  of  finding  the  centre  of 
gravity  is  to  carefully  balance  the  template  upon  a  needle-point. 

The  area  of  such  a  section  may  be  determined  either  by 
means  of  a  planimeter  or  by  balancing  the  template  against  a 
rectangle  cut  out  of  the  same  material,  the  area  of  the  rectangle 
being  evidently  the  same  as  that  of  the  section. 

7.  Moment  of  Inertia  of  a  Plane  Area. — Let  any  two 
consecutive  sides,  C^C^,  C^C^■,o[  the  funicular  polygon  meet 
line  fG  in  the  points  m^ ,  n^ . 

Let  :ir, ,  4:, ,  .Tj ,  .  .  .  be  the  lengths  of  the  perpendiculars  from 
thecentrcsof  gravity  of  yj,,  A^,  A^,  .  .  .  ,  respectively,  upon  ^G^. 

Draw  the  line  OH  perpendicular  to  the  line  of  loads,  and 
let  OH  =p. 

By  the  similar  triangles  C^vi^n^  and  d?34, 


X, 


34       «,  «3^, 

P      P  p 


lii  % 


P   2" 


m^n^—  =  area  of  triangle  C,w,ji,, 


But  the  total  area  A  bounded  by  the  funicular  polygon 
C,CjC,  .  .  .  and  the  lines  ^Ci ,  gk  is  the  sum  of  all  the  triangular 
areas  C,gm^,  C,w/,;/, ,  C^ifi^n^,  .  .  .,  described  in  the  same  manner 
as  C\m,n,. 


a,x. 


/>    2     '    /    2 


+  ...= 


2/     " 


The  sum  2{ax'')  is  the  moment  of  inertia,  /,  of  the  plane 
area  with  respect  to  gG.     Hence, 


A  =—.     or    /  =  2Ap, 
2p 


h  ' 


MOMENT  OF  INERTIA. 


n 


The  moment  of  inertia  /^  of  the  area,  with   respect  to  a 
parallel  axis  at  distance  ^^  from  gG,  is  given  by  the  equation 

\vhere5=  ^,  +  /4,  +  .  .  . 

I  et  the  new  axis  intersect  Cyg  and  kg  in  the  points  q  and  r. 
The  triangles  qgr  and  Om  are  similar. 

qr      I «       S 

"  y~  P  ~  /  ' 


and,  therefore,  the  area  A'  of  the  triangle  qgr 


'Im 


Hence 


ly  =  2pA  -\-2pA'  =  2p{A  4-  A'). 


I W  y4 

iVi?^^. — If  />  be  made  =  —  =  — , 

2  2 


and 


7=^'     and      5y,'rr^^', 


.-.  /,  =  .4(/i+^';- 


The  angle  lOn  is  also  evidently  a  right  angle. 

8.  Cranes. — (a).  Jib-crane. — Fig.  19  is  a  skeleton  diagram  of 
an  ordinary  jib-crane.  OA  is  the  post  fixed  in  the  ground  at 
0\  OB  is  the  jib  ;  AB  is  the  tie.  The  jib,  tie,  and  gearing  are 
suspended  from  the  top  of  the  post  by  a  cross-head,  which 
admits  of  a  free  rotation  round  the  axis  of  the  post. 

Let  the  crane  lift  a  weight  W. 


:l 


|i|  t  in 


14 


THEORY  OF  STRUCTURES. 


Three  forces  in  equilibrium  meet  at  B\  viz.,  W,  the  tension 
T  in  the  tie,  and  the  thrust  C  along  the  jib. 


Fig.  ao. 


Draw  the  reciprocal  figure  55,5,  of  B,  S^S^  representing  W. 


T       SS„       AB 


and 


W~  S,S,~  Aa 
W~  S,S,'~  Ad' 


The  load  is  not  suspended  directly  from  ^,  but  is  carried  by 

a  chain  passing  over  pulleys  to  a  chain-barrel  usually  fixed  to 

the  crane-post.     The  stress  5  in  the  chain  depends  upon  the 

W 
system  of  pulleys,  and  is,  e.g.,  — ,  if  «  is  the  number  of  falls  of 

chain  from  B  and  if  friction  is  7teglected.  In  order  to  obtain 
the  true  values  of  7"and  6"  this  tension'5must  be  compounded 
with  W. 

Draw  5,/t  parallel  to  the  direction  in  which  the  chain  passes 
'  om  B  to  the  chain-barrel,  and    take  S^k  to  represent  5   in 


CRANES. 


15 


magnitude.  The  line  S,k  evidently  represents  the  resultant 
force  at  B  due  to  IV  and  S. 

Draw  ki  parallel  to  AB. 

The  tension  in  the  tie  and  the  thrust  in  the  jib  are  now 
evidently  represented  by  tk,  tSi ,  respectively. 

Generally  the  effect  of  chain-tension  is  to  ditiiinish  the  ten- 
sion of  the  tie  and  to  increase  the  thrust  on  the  jib. 

,  ^     .       ^BD        „.BD 
Thevertical  component  of  T,  viz.,  B—rj; 

mitted  through  the  post. 


\V- 


AO' 


is  trans- 


Thc  total  resultant  pressure  along  the  post  at  O 


Tsin  BAB -\-C  sin  BOF=-lV^^-\-lV-^ — -J^ 


""^    ivli^^^±^=jv. 


The  pull  upon  the  tie  tends  to  upset  the  crane,  and  its 
moment  with  respect  to  O  is 

r  cos  BAD  XA0=  W^^AO  =  WAD  =  WOF, 

AU  AB 

OF  being  the  horizontal  projection  of  AB. 

OF  is  often  called  the  radius  or  throw  of  the  crane. 

If  the  post  revolves  about  its  axis  (as  in //V-cranes),  the  jib 
and  gearing  are  bolted  to  it,  and  the  whole  turns  on  a  pivot  at 
the  toe  G.  In  this  case,  the  frame,  as  a  whole,  is  kept  in 
equilibrium  by  the  weight  W,  the  horizontal  reaction  //  of  the 
web-plate  at  O,  and  the  reaction  R  at  G.  The  first  two  forces 
meet  in  F  and,  therefore,  the  reaction  at  G  must  also  pass 
through  F. 

Hence,  since  OFG  may  be  taken  to  represent  the  triangle 
of  forces. 


H=  W 


OF 
OG 


and     R  =  W- 


GF 


OG' 


In  a  portable  crane  the  tendency  to  upset  is  counteracted 
by  means  of  a  weight  0  placed  upon  a  horizontal  platform 
OL  attached  to  the  post  and  supported  by  the  tie  AL. 

The  horizontal  projection  fm  of  fk  represents  the  horizontal 


i  'i  i 


i6 


THEORY  OF  STRUCTURES. 


Fig.  31. 


pull  at  A,  and  if  tn  be  drawn  parallel  to  AL,  the  intercept  vin 
cut  off  on  the  vertical  through  m  by  the  lines  tin  and  tn  repre- 
sents the  counter-weight  required  at  L. 

{b)  Derrick-crane, — The  figure  shows  a  combination  of  a  der- 

B  rick  and  crane,  called  a  derrick- 
crane.  It  is  distinguished  from 
the  jib-crane  by  having  two 
back-stays,  AD,  AE.  One  end 
of  the  jib  is  hinged  at  or  near 
the  foot  of  the  post,  and  the 
other  is  held  by  a  chain  which 
passes  over  pulleys  to  a  winch 
on  the  post,  so  that  the  jib  may 
be  raised  or  lowered  as  required. 
The  derrick-crane  is  gener- 
ally of  wood,  is  simple  in  con- 
struction, is  easily  erected,  has 
a  vertical  as  well  as  a  lateral 
motion,  and  a  range  equal  to  a  circle  of  from  lo  to  60  feet 
radius.  It  is  therefore  useful  for  temporary  works,  setting 
masonry,  etc. 

The  stresses  in  the  jib  and  tie  are  calculated  as  in  the  jib- 
crane,  and  those  in  the  back-stays  and  post  may  be  obtained  as 
follows  : 

Let  the  plane  of  the  tie  and  jib  intersect  the  plane  DAE  of 
the  two  back-stays  in  the  line  AF,  and  suppose  the  back-stays 
replaced  by  a  single  tie  AF.  Take  OF  to  represent  the  hori- 
zontal pull  at  A.  The  pull  on  the  "  imaginary"  stay  A F  is  then 
represented  by  AF  and  is  evidently  the  resultant  pull  on  the 
two  back-stays.  Completing  the  parallelogram  EG  AH,  AH 
will  represent  the  pull  on  the  back-stay  AF.,  and  AG  that  upon 
AD,  their  horizontal  components  being  OK,  OL,  respectively. 
The  figure  OKFL  is  also  a  parallelogram. 

If  the  back-stays  lie  in  planes  at  right  angles  to  each  other, 

OL  =  OF  cos  B  —  T  sin  a  cos  Q,  and  is  a  max.  when  ^  =    0°, 

and 


OK  =^  OF  sin  6  =  T  sin  <>e  sin  6,  and  is  a  max.  when  6 


90°, 


BRIDGE   AND  ROOF   TRUSSES. 


17 


B  being  the  angle  FOL,  and  a  the  inclination  of  the  tie  to  the 
vertical. 

Hence  the  stress  in  a  back-stay  is  a  maximum  when  the 
plane  of  the  back-stay  and  post  coincides  with  that  of  the  jib 
and  tie. 

Again,  let  (i  be  the  inclination  of  the  back-slays  lu  Lhe  ver- 
tical.    The  vertical  components  of  the  back-stay  stresses  are 

Tsin  a  cos  6*  cot  ft    and     T  sin  n  sin  (9  cot  ft\ 

and,  therefore,  the  corresponding  stress  along  the  post  is 

T  sin  a  cot  /i  (cos  B  -\-  sin  if), 

which  is  a  maximum  when  B  =  45°. 

9.  Shear  Legs  (or  Shears)  and  Tripods  (or  Gns)  are 


ow 


Pig.  23. 


often  employed  when  heavy  weights  are  to  be  lifted.  The 
former  consists  of  two  struts,  AD,  AE.  united  at  A  and  sup- 
ported by  a  tie  A  C,  which  may  be  made  adjustable  so  as  to 
admit  of  being  lengthened  or  shortened.  The  \vei<dit  is  sus- 
pended from  A,  and  the  legs  are  capable  of  revolving  around 
DE  as  an  axis.  Let  the  plane  of  the  tie  and  weight  intersect 
the  plane  of  the  legs  in  AF,  and  suppose  the  two  legs  replaced 
by  a  single  strut  AF.  The  thrust  along  AF  can  now  be 
easily  obtained,  and  hence  its  components  along  the  two  legs. 

In  tripods  one  of  the  three  legs  is  usually  longer  than  the 
others.  They  are  united  at  the  top,  to  which  point  the  tackle 
is  also  attached. 

10.  Bridge  and  Roof  Trusses  of  Small  Span.— A  single 
girder  is  the  simplest  kind  of  bridge,  but  is  only  suitable  fur 


tS 


THEORY  OF  STRUCTURES. 


very  short  spans.  When  the  spans  are  wider,  the  centre  of  the 
girder  may  be  supported  by  struts  OC,  OD,  through  which  a 
portion  of  the  weight  is  transmitted  to  the  abutments. 


\ 

0 

B 

'/A'A 

>^\^ 

i" 

1 

y 

-^^pX 

.    t 

^ 

•; 

y^ 

\.  ^ 

i 

To 

Fig. 


as- 


Fig.  24. 


Take  the  vertical  line  5,5",  to  represent  P,  the  weight  at  O. 
Draw  55i  parallel  to  OC,  and  SS^  to  OD. 

Draw  the  horizontal  SH,  and  let  the  angle  AOC  =:^  a. 


The  thrust  along  OC  =  S^S  =  S^H  cosec  a  =  —  cosec  a. 

P 
The  tension  along  OA  =  SH  =  S,H  cot  a  =  —  cot  a. 

The  horizontal  and  vertical  thrusts  upon  the  masonry  at  C 

P  P 

(or  D)  are  —  cot  a  and  — ,  respectively. 

If  the  girder  is  uniformly  loaded,  P  is  one  half  of  the  whole 
load. 

II.  In  the  figure  a  straining  cill,  EF,  is  introduced,  and  the 
girder  is  supported  at  two  intermediate  points. 


1 
f 


Fig.  as. 


Fig.  s6. 


Let  Pht  the  weight  at  each  of  the  points  E  and  F. 
Draw  the  reciprocal  SS,Ho[  the  point  E,  ^i//" representing  P. 


ill 


BRIDGE  AND  ROOF  TRUSSES. 


19 


The  thrust  in  EC  (or  FD)  =  SS,  =  p:^%  =  P^.  and  the 

o,/7  AL 


horizontal  thrust  in  the  straining  piece 


If  a  load  is  uniformly  distributed  over  AB,  it  may  be 
assumed  that  each  strut  carries  one  half  of  the  load  upon  AF 
(or  BE),  and  that  each  abutment  carries  one  half  of  the  load 
upon  AE  (or  BF). 

By  means  of  straining  cills  the  girders  may  be  supported  at 
several  points,  i,  2,  .  .  .  ,  and  the 
weight  concentrated  at  each  may 
be  assumed  to  be  one  half  of  the 
load  between  the  two  adjacent 
points  of  support.  The  calcula- 
tions for  the  stresses  in  the  struts, 
etc.,  are  made  precisely  as  above. 

If  the  struts  are  very  long  they  are  liable  to  bend,  and 
counterbraces,  AM,  BN,  are  added  to  counteract  this  tendency. 

12.  The  triangle  is  the  only  geometrical  figure  of  which 
the  form  cannot  be  changed  without  varying  the  lengths  of  the 
sides.  For  this  reason,  all  compound  trusses  for  bridges,  roofs, 
etc.,  are  made  up  of  triangular  frames. 

Fig.  28  represents  the  simplest  form  of  roof-truss.  AC, 
BC  a.rQ  rafters  of  equal  length  inclined  to  the  horizontal  at  an 
angle  01,  and  each  carries  a  uniformly  distributed  load  iV. 


Fig.  a?. 


H 


■  i 


> .  \i 


,11 


Fig.  28. 


The  rafters  react  horizontally  upon  each  other  at  C,  and 
their  feet  are  kept  in  position  by  the  tie-beam  AB.  Consider 
the  rafter  A  C. 

The  resultant  of  the  load  upon  AC,  i.e.,  IF,  acts  through 
the  middle  point  D. 


I 


90  THEORY  OF  STRUCTURES. 

Let  it  meet  the  horizontal  thrust  //  of  BC  upon  AC  \t\  F. 
For  equilibrium,  the  resultant  thrust  at  A  must  also  act 
through  F. 

The  sides  of  the  triangle  AFE  evidently  represent  the 
three  forces.     Hence 


„     ,,,AE      WAR      W     , 

"—  ^iTr-  —  —  77P  =  — cot  a ; 
hb        2  DE       2' 


R  =  wf^=W. 


-\-£F' 


V     ^^ 


=  V-^i(^2'=V-  + 


cot  a 


The  thrust  R  produces  a  tension  H  in  the  tie-beam,  and  a 
vertical  pressure  W  upon  the  support. 
Also,  if  y  is  the  angle  FAE, 


FF         DF 


^ 


If  the  rafters  AC,  BC  are  unequal,  let  or,,  or,  be  their  in- 
clinations to  A,  B,  respectively. 

Let  W^  be  the  uniformly  distributed  load  upon  AC,  W, 
that  upon  BC. 

Px^M F/'^i 


Fic.  39. 

Let  the  direction  of  the  mutual  thrust  P  zt  C  make  an 
angle  /?  with  the  vertical,  so  that  if  CO  is  drawn  perpendicular 


ROOF   TRUSSES. 


ai 


to  i^C,  the   angle   COB  =  fi\  the   angle    ACF=go°  -  ACO 

=  QO'^  -  1^  -  <^). 

Draw  AM  perpendicular  to  the  direction  of  P,  and  consider 
the  rafter  AC.  As  before,  the  thrust  A',  at  A,  the  resultant 
weight  lV^  at  the  middle  point  of  AC,  and  the  thrust  P  aX.  C 
mecL  ill  the  point  F. 

Take  moments  about  A.     Then 

P.AM=  W,AE. 
But  AM  =  AC  sin  A  CM  =  AC  cos  {/3  —  or,), 


and         ^£  =  — -  cos  or,. 

4W 


P  = 


^^. 


cos  o'. 


2  cos(/^  —  ar,)' 
Similarly,  by  considering  the  rafter  BC, 


P  = 


W., 


cos  a„ 


W. 


cos  or. 


2  sin(/S  +  «,  —  90°) 


2  cos(/S4-«,)' 


Hence 


cos  a, 


w. 


2  cos{fi  —  a,) 


cos  a„ 


2   cosi/H -\- a;)' 


and  therefore 


tan  /?  = 


PF,+  ^, 


W^  tan  «,  —  W^3  tan  a 


The  horizontal  thrust  of  each  rafter  =  Psin  /?. 

The  vertical  thrust  upon  the  support  A  =  W^  —  P  cos  fi.  » 

The  vertical  thrust  upon  the  support  B  =  W^-\-  P  cos  /?. 

13.  King-post  Truss. — The  simple  triangular  truss  may 
be  modified   by    introducing  a  C 

king-post  CO,  which  carries  a 
portion  of  the  weight  of  the 
beam  AB,  and  transfers  it 
through  the  rafters  so  as  to  act  p^^ 

upon  the  tie  in  the  form  of  a  tensile  stress. 


w 


■Til 


I;.:   ;i 


a2 


THEORY  OF  STRUCTURES. 


Let  P  be  the  weight  borne  by  the  king-post ;  represent  it 
by  CO. 

Draw  OD  parallel  to  BC,  and  DE  parallel  to  AB. 


DC  = 


CE 
sin  u 


cosec  a  is  the  thrust  in   CA  due  to  P,  and 


is  of  course  equal  to  DO,  i.e.,  the  thrust  along  CB. 

P 
DE  =  CE  cot  01  =  —  cot  a   is    the   horizontal     thrust    on 

each  rafter,  and  is  also  the  tension  in  the  tie  due  to  P. 

Let  W  be  the  uniformly  distributed  load  upon  each  rafter. 

cot  ^f 


The  total  horizontal  thrust  upon  each  rafter  =  ( W-\-  P)- 
The  total  vertical  pressure  upon  each  support  =  W  -\- 


P 


If  the  apex  C  is  not  vertically  over  the  centre  of  the  tic- 
beam  take  COy  as  before,  to  represent  the  weight  P  borne  by 

the  king-post ;  draw  (^Z)  parallel 
to  BC,  and  DE  parallel  to  AB. 
The    weight   P  produces   a 
B       thrust  CD  along  CA,  DO  along 
fe^S    CB,  and  a  horizontal  thrust  DE 
^'^^-  3i-  upon  each  rafter. 

CE  is  the  portion  of  P  supported  at  A,  and  EG  that  sup- 
ported at  B. 

DE,  and  therefore  the  tension  in  the  tie  AB,  diminishes 

with  AO,  being  zero  when  AC 
is  Vertical. 

Sometimes  it  is  expedient 
to  support  the  centre  of  the  tie- 
beam  upon  a  column  or  wall, 
the  king-post  being  a  pillar 
against  which  the  heads  of  the 
rafters  rest. 

Consider  the  rafter  AC. 
The  normal    reaction  R'  of 
CO    upon    AC,    the    resultant 
KiG,  3a.  weight  W  at  the  middle  point 

D,  and  the  thrust  R  dX  A  meet  in  the  point  F. 


In ' 


ROOF    TRUSSES. 
lake  moments  about  A.     Then 


n 


w 


R'AC=  W.AE,     or    R' =  —  cos  a. 


Thus  the  total  thrust  transmitted  through  CO  to  the  sup- 

W 
port  at  0  is  2 —  cos  oe .  cos  a  =  W  cos'  a. 

The  horizontal  thrust  upon  each  rafter 

W  .  W    . 

=  —  cos  a  sm  a  =  —  sm  2(x. 

2  4 

14.  If  the  rafters  are  inconveniently  long,  or  if  they  are  in 
danger  of  bending  or  breaking  transversely,  the  centres  may 
be  supported  by  struts  OD,  OE.     A  portion  of  the  weight  upon 


Fio. 


34- 


the  rafters  is  then  transmitted  through  the  struts  to  the 
vertical  tie  (king-post  cr  rod)  CO,  which  again  transmits  it 
through  the  rafters  to  act  partly  as  a  vertical  pressure  upon 
the  supports,  and  partly  as  a  tension  on  the  tie-beam.  The 
main  duty,  indeed,  of  struts  and  tics  is  to  transform  transverse 
into  longitudinal  stresses. 

This  king-post  truss  is  the  simplest  and  most  economical 
frame  for  spans  of  less  than  thirty  feet.  In  larger  spans  two 
or  more  suspenders  may  be  introduced,  or  the  truss  otherwise 
modified. 


mi: 


l|! 


H 


THEORY  OF  STRUCTURES. 


%  % 


iHii 


Let  there  be  a  load  2  W  uniformly  distributed  over  the 
rafters  ACy  BC,  and  assume  it  to  be  concentrated  at  the  joints 

A,  D,  C,  E,  B,  m  the  proportion  — ,  — ■,  — ,  — ,  — . 

42224 

Also,  let  the  load  (including  a  pol-tion  of  the  weight  upon 

the  tie-beam  AB,  and  the  weights  of  the  members  OD,  OE, 

OC)  borne  directly  at  O  be  P. 

P 
The  total  reaction  at  each  support;  is  W -\ — ,  and  acts  in 

W    ' 
an  opposite  direction  to  the  weight  —  there  concentrated. 

4 

Hence  the  resultant  reaction  at  a  support  is  ^W-\-\P. 
Thus,  the  weights  at  the  points  of  support  A  and  B  are 
taken  up  by  the  abutments,  and  need  not  be  considered  in  de- 
termining the  stresses  in  the  several  members  of  the  frame. 

Draw  the  reciprocal  S^HS^  of  ^.    Then 

xW     P 

SyH  = 1 —  ;    //"i^,  =  tension  in  AO; 

42 

5g5i  =  compression  in  AD. 
Draw  the  reciprocal  S^S^S3S^  of  D.    Then 

5,5,  =  compression  in  OB;    S^S^  =  compression  in  DC; 


5,5,  =  —  =  weight  at  D. 

2 


Draw  the  reciprocal  S^S^S^S^S^  of  C,   Then 


5,5,  =  tension  in  CO  =  —  -f-  /*;   5,5,  =  compression  in  CE; 


W 

S^S^  =  —•  —  weight  at  C. 


CE\ 


ROOF    TRUSSES. 

Draw  the  reciprocal  5,5,5,5,5^  of  E.    Then 


35 


S^S^  =  compression  in  OE ;    5,5,  =  compression  in  BE; 

W 


0,0, 


=  weight  at  E. 


Draw  S^K  horizontally.    Then 

S,KS\S,  is  evidently  the  reciprocal  of  B ;  A'5,  =  f  W-\-  ^P, 
being  \he  reaction  at  B,  and  S^K  the  tension  in  the  tie  BO.  The 
reciprocal  of  O  is  also  the  figure  SjHKStS^S,S^ ,  and  //A!"  =:  P. 

15.  Collar-beams  (/?£■),  queen-posts  (DF,  EG),  braces,  etc., 
may  bo  employed  to  prevent  the  deflection  of  the  rafters. 
The  complexity  of  the  truss  necessarily  increases  with  the  span 
and  with  the  weight  to  be  borne. 


A'-" sr^^ 


Pig.  35 


Fig.  36. 


With  a  single  collar-beam  and  a  uniformly  distributed  load, 
SJ7S,  is  the  reciprocal  of  A,  and  5,5,^5,5,5,  the  reciprocal  of 
B ;  5,//  being  the  reaction  at  A,  and  5^5,  the  weight  at  B. 


D 

/^\ 

E 

\^ 

^\B 

t\ 

-            c 

^          U- 

Fig.  37. 


Pig.  38, 


With  a  collar-beam  DE,  two  king-posts  DF,  EG,  and   a 
uniformly  xlistributed  load,  the  stresses  at  the  joints  D  and  E 


1 

in 

>"* 

r 

!ii[ 

1 

I  ^i 


26 


THEORY  OF  STRUCTURES. 


become  indeterminate.  To  render  them  determinate  it  is 
sometimes  assumed  that  the  components  of  the  weights  at  D 
and  E,  normal  to  the  rafters,  are  taken  up  by  the  collar-beam 
and  corresponding  king-post.  Thus  S^HS^  is  the  reciprocal  of 
A,  and  ^,5,535<5j  the  reciprocal  of  D,  S^H  being  the  reaction  at 
Ay  6"j,.S,  the  weight  at  D\  SJS.,  is  the  normal  component  of  the 
weight,  and  the  components  of  S^S^,  viz.,  SJS^  horizontal  and 
Sj^.,  vertical,  represent  the  stresses  borne  by  DE  and  DP,  re- 
spectively. 

This  frame  belongs  to  the  incomplete  (Art.  1 8)  class,  and  if 
it  has  to  support  an  unequally  distributed  load,  braces  must  be 
introduced  from  D  to  G  and  from  E  to  F. 

l6.  The  truss  ABC,  Fig.  40,  having  the  rafters  supported 
at  two  intermediate  points,  maybe  employed  for  spans  of  from 
30  to  50  feet.  Suppose  that  these  intermediate  points  of  sup- 
port trisect  the  rafters,  and  let  each  mfter  carry  a  uniformly  dis- 
tributed load  W. 


Fig. 


39- 


Fig.  40. 


Then  a  weight  may  be  considered  as  concentrated  at  eacli 

W 
of  the  joints  //,  D,  C,  E,  K.     This  wei'dit  =  — . 

3 
Let  P  be  the  weight  directly  supported   at  each   of  the 
joints  F,  G. 

The  resultant  reaction  Vit  A  —  l\V-\-  P. 

S^//S\,  is  the  reciprocal  of  A,  S^H  representing  ^W-{-  P. 

S^SjS^S^S^  is  the  reciprocal  of  //. 

S^SJ/KS,S,  is  the  reciprocal  of  P,  HK  representing  P,  the 

weight  directly  borne  at  P. 
6^,5,5,5,5,53  is  the  reciprocal  of  />,  S^S,  representing  the 

weight  at  D,  S,S^  the  thrust  along  HD,  S,S,  the 

tension  in  DP,  S,S,  the  thrust  along  El),  and 

S^S,  the  thrust  along  CV. 


INCOMPLETE  FRAMES. 


V 


;  each 


)f  the 


5.  the 


As  in  the  preceding  case,  this  truss  will  be  found  incomplete 
if  the  load  is  unevenly  distributed,  and  the  reciprocals  of  D  and 
E  will  not  close.  In  practice,  however,  the  friction  at  the  joints, 
the  stiffness  of  the  several  members,  and  the  mode  of  construc- 
tion render  the  truss  sufificiently  strong  to  meet  the  ordinary 
variations  of  load. 

17.  General  Remarks. — In  the  trusses  described  in  Arts. 
n  and  14  the  vertical  members  are  ties,  i.e.,  are  in  tension,  and 
the  inclined  members  are  struts,  i.e.,  are  in  compression.  By 
inverting  the  respective  figures  another  type  of  truss  is  obtained 
in  which  the  verticals  are  struts  while  the  inclined  members  are 
ties.  Both  systems  are  widely  used,  and  the  method  of  calcu- 
lating the  stresses  is  precisely  the  same  in  each. 

In  designing  any  particular  member,  allowance  must  be 
made  for  every  kind  of  stress  to  which  it  may  be  subjected. 
The  collar-beam  DE,  for  example,  must  be  treated  as  a  pillar 
subjected  to  a  thrust  in  the  direction  of  its  length  at  each  end  ; 
if  it  carry  a  transverse  load,  its  strength  as  a  beam,  supported 
at  the  points  D  and  Ey  must  also  be  determined.  Similarly, 
the  rafters  AC,  BC,  etc.,  must  be  designed  to  carry  transverse 
loads  and  to  act  as  pillars.  But  it  must  be  remembered  that 
struts  and  queen-posts  provide  additional  points  of  support 
over  which  the  rafters  are  continuous,  and  it  is  practically  suf- 
ficient to  assume  thart  the  rafters  are  divided  into  a  number  of 
short  lengths,  each  of  which  carries  one  half  oi  the  load  between 
the  two  adjacent  supports. 

When  a  tie-beam  is  so  long  as  to  require  to  be  spliced, 
allowance  must  be  made  for  the  weakening  effect  of  the  splice. 
18.  Incomplete  Frames. — The  frames  discussed  in  the 
preceding  articles  (excepting  those  referred  to  in  Art.  15)  will 
support,  without  change  of  form,  any  load  consistent  with 
strength,  and  the  stresses  in  the  several  members  can  be  found 
in  terms  of  the  load.  It  sometimes  happens,  however,  that  a 
frame  is  incomplete,  so  that  it  tends  to  change  form  under  every 
distribution  of  load.  An  example  of  this  class  is  the  simple 
trapezoidal  truss,  consisting  of  the  two  horizontal  members  AB, 
DE,  and  the  two  equal  inclined  members  AD,  JUi,  Fig.  41. 
First,  let  there  be  a  weight  W  at  each  of  the  points  D,  E. 


1  , 


ill'-! 


38 


THEORY  OF  STRUCTURES. 


The  triangles  of  forces  for  the  joints  D  and  E,  viz.,  SS^H  and 
SSyH,  can  be  drawn,  and  hence  it  follows  that  there  must  be 


Fig.  41. 


Fig.  42. 


equilibrium.     This  is  also  evident  from  the  symmetrical  char- 
acter of  the  loading. 

The  same  triangles  represent  the  forces  at  the  points  of 
support  A,  B. 

.-.  reaction  at  .r^  —  S,I/=  W=  S^H  =  reaction  at  B. 


Next,  let  iherc     e 
<<  W,)  at  E. 


\;«;ight  fF,  at  D  and  a  weight   W, 


Fig.  43. 

It  will  now  be  found  that  the  diagram  of  forces  will  not 
close,  so  that  there  cannot  be  equilibrium.  The  joint  B  will  be 
pushed  in  and  the  frame  distorted.  The  distortion  may  be 
prevented  by  introducing  a  brace  irom  A  to  E  or  from  B  to  D. 
In  the  latter  case  S^mSS^S^  represents  the  stress-diagram,  the 
triangle  S^HS  being  the  reciprocal  of  the  joint  E,  and  the  quad- 
rilateral Sm S ^ f/  thiit  of  the  joint  D.  Drawing  the  horizontal 
;«//,  the  triangle  mnS^  and  the  quadrilateral  mSS^n  are  evidently 
the  reciprocals  of  A  and  B,  respectively. 


.*.  nS\  =  reaction  at  A     and     «S,  =  reaction  at  B. 


INCOMPLE  TE  FRA  MES. 


29 


In  practice  the  loads  are  usually  transmitted  toZ?  and  E  by 
means  of  two  vertical  queenposis  {queen-rods  or  queens)  DF,  EG. 


I 

D               1 

^ 

A/ 

\ 

\b 

1 

=               G 

Fig.  45. 


If  there  are  no  diagonal  braces  DG,  EF,  the  distortion  of 
the  frame  under  an  unevenly  distributed  load  can  only  be  pre- 
vented by  the  friction  at  the  joints,  the  stiffness  of  the  mem- 
bers, and  by  the  queens  being  rigidly  fixed  to  AB  at  /'"and  G. 

Let  W^  be  the  load  at  F  transmitted  through  the  queen 
FD  to  D. 

Let  W^  (<  W^  be  the  load  at  G  transmitted  through  the 
queen  GE  to  E. 

If  the  frame  is  rigid,  the  reactions  R^  at  A  and  R^  at  B, 
which  will  balance  these  weights,  can  easily  be  found  by  taking 
moments  about  B  and  A,  successively.     Thus, 


and 


W  W 


W  IV 

RJ  = -:r{l  -  c)  + -^{l -V  c), 


where  AB  =  I  and  FG  =  c. 

Draw  the  triangle  of  forces  SHS^  for  the  joint  A,  5//"  rep- 
resenting R^ . 

The  triangle  SS,X  is  the  reciprocal  of  the  joint  at  D,  and  the 
tension  in  FD  should,  therefore,  be  XS^  =  SH  =.  7?, .      But  the 


';ii 


M 


M  r 


I 


1  i    ' 


i:^l 


3© 


THEORY  OF  STRUCTURES 


tension  in  FD  is  actually  W^ ,  so  that  there  is  an  unbalanced 
force, 


=  W,  -  R, 


W,  -  W,    l-c 

2  '       /     ' 


acting  along  FD. 

To  take  up  this  unbalanced  force  and  render  the  frame  rigid 
the  diagonal  DG  is  introduced,  and  the  stress  for  which  it 
should  be  designed  is  evidently 

W  —  W  I  —  c  s 
( W,  -  R^)  sec  FDG  =  --^ ^'  '—^  - , 

s  being  the  length  of  the  diagonal  and  a?  the  depth  of  the  truss. 
The  complete  stress  diagram  is  as  shown  in  Fig.  46. 

Cor.  I.  The  manner  in  which  distortion  is  prevented  by  the 
stiffness  of  AB  may  be  shown  as  follows: 

Let  X  be  the  force  of  resistance  which  AB,  by  its  stiffness, 
can  exert  at  F  or  6^  against  any  load  which  tends  to  make  it 
deviate  from  the  horizontal. 

If  W\s  the  load  at  F,  the  actual  downward  pull  upon  D  is 
W—  X ;  this  must  necessarily  produce  an  equal  upward  pull  at 
E,  which  must  be  balanced  by  the  force  of  resistance  x  at  G, 


'.     W  —  X  =  X, 


and 


W 


X  = 


'i 
1:: 

I 

i 

Thus  the  beam  AB  will  be  acted   upon  by  an  upward  pull 

W 

—  at  F  and  an  equal  downward  pull  at  G,  forming  a  couple 

IV 
of  moment  — c,  and  showing  that  equilibrium  is  impossible. 

The  upward  reaction  R^  at  A  is 

'  ~"  7v  T  ~2         ~  "2        2~"/  ~  ~J1 


I 


COMPOSITE  FRAMES.  %\ 

=  downward  reaction  at  B^  and  the  moment  at  F  (or  G) 


2/2 


Cor.  2.  Let  a  weight  IF  be  supported  at  the  joint  D  of  anj 
quadrilateral  frame  ADEB.     Draw  the  reciprocal  SS^S^  of  D, 


Fig.  47- 


Fig.  48. 


5,5,  representing  W.  Draw  ^S,  parallel  to  EB  and  intersecting 
the  vertical  5,5,  produced  in  53 .  The  weight  which  can  be 
borne  at  E  consi.stent  with  equilibrium  is  represented  b}'  5,53 . 

19.  Composite  Frames  or  Trusses  (i.e.,  frames  made  up 
of  two  or  more  simple  frames). — An  example  of  this  class  has 
already  been  given  in  the  case  of  the  king-post  roof  (Art.  13). 

Bent  Crane. — Fig.  49  shows  a  convenient  form  of  crane 
when  much  head-room  is  required  near  the  post.  The  crane 
is  merely  a  semi-girder,  and  may  be  tubular  with  plate-webs  if 
the  loads  are  heavy,  or  its  flanges  may  be  braced  together  as 
in  the  figure  for  loads  of  less  than  ten  tons.  The  flanges  may 
be  kept  at  the  same  distance  apart  throughout,  or  the  distance 
may  be  gradually  diminished  from  the  base  towards  the  peak. 

Let  the  nuqpbers  in  Fig.  50  denote  the  stresses  in  the  cor- 
responding members.  Three  forces,  5, ,  C, ,  and  W,  act  through 
the  point  (l),  so  that  5,  and  (7,  may  be  obtained  in  terms  of 
W\  three  forces,  S^,  S^,  T,,  act  through  (2),  so  that  5,  and  7", 
may  be  obtained  in  terms  of  5,  and  therefore  of  IV;  four 
forces,  5j,  (T,,  5,,  (7<,  act  through  (3),  and  the  values  of  5, ,  Ci 


32 


THEORY  OF  STRUCTURES. 


!  1 


being  known,  those  of  vS,,  C^  may  be  determined.  Proceed- 
ing in  this  way,  it  is  found  that  of  the  forces  at  each  succeed- 
ing joint  only  two  are  unknown,  and  the  values  of  these  are 
consequently  determinate. 


Fio.  so. 

The  calculations  may  be  checked  by  the  method  of  mo7nents. 
and  by  the  stress  diagram  (Fig.  50). 
E.g.,  let  W=  10  tons. 
T?  :e  moments  about  the  point  (7).    Then 

r,(.r7)  =  10(^7)  or  r,  =  ^  =  26  tons  =  (68)  in  Fig.  50. 

I. -5 

No  other  forces  enter- into  the  equation  of  moments,  as  the 
portion  of  the  crane  above  a  plane  intersecting  (68)  and  passing 
through  (7)  is  kept  in  equilibrium  by  the  weight  of  10  tons 
and  the  stresses  T^,  S^,  C^;  the  moments  of  5,  and  C,  about 
(7)  are  evidently  zero. 

In  the  stress  diagram  (Fig.  50)  PaQ  is  the  reciprocal  of  the 
point  I,  adQ  of  the  point  2,  PcdQ  of  3,  Qdai  q^  4,  and  so  on. 

Other  examples  of  composite  roof  and  bridge  frames  will 
now  be  given. 

20.  Roof-trusses. — A  roof  consists  of  a  covering  and  of 
trusses  (or  frames)  by  which  it  is  supported.  The  covering 
is  generally  laid  upon  a  number  of  common  rafters  which  rest. 


ROOF   TRUSSES. 


33 


upon  horizontal  beams  (or  purlins),  the  latter  being  carried 
y-y  trusses  spaced  at  intervals  varying  with  the  type  of  con- 
struction but  averaging  about  lO  ft.  The  truss  rafters  arc 
called  principal  rafters,  and  the  trusses  themselves  are  often 
designated  as  principals. 

In  roofs  of  small  span  the  trusses  and  purlins  are  sometimes 
dispensed  with. 

Types  of  Trtiss. — A  roof-truss  may  be  constructed  of  tim- 
ber, of  iron  or  steel,  or  of  these  materials  combined.  Timber 
is  almost  invariably  employed  for  small  spans,  but  in  the 
longer  spans  it  has  been  largely  superseded  by  iron,  in  con- 
sequence of  the  combined  lightness,  strength,  and  durability 
of  the  latter. 

Attempts  have  been  made  to  classify  roofs  according  to  the 
mode  of  construction,  but  the  variety  of  form  is  so  great  as  to 
render  it  impracticable  to  make  any  further  distinction  than 
that  which  may  be  drawn  between  those  in  which  the  reac- 
tions of  the  supports  are  vertical  and  those  in  which  they  are 
inclined. 


Fig.  51. 


Fig.  S3. 


Fig.  S3. 


Fig.  54. 


Fig.  ss. 


Fig.  s6. 


Fig.  57. 


Fig.  58. 


11; 
'(,[1? 


Fig.  59. 


\\ 


Fig.  51  is  a  simple  form  of  truss  for  spans  of  less  that  30  ft. 

Fig.  52  is  a  superior  framing  for  spans  of  from  30  to  40  ft.; 
it  may  be  still  further  strengthened  by  the  introduction  of 
struts,  Figs.  53  and  54,  and  with  such  modification  has  been 
employed  to  span  openings  of  90  ft.  It  is  safer,  however,  to 
limit  the  use  of  the  type  shown  by  Fig.  53  to  spans  of  less  than 


It 


34 


THEORY  OF  STRUCTURES. 


60  ft.  Figs.  55,  56,  57,  58,  and  59  are  forms  of  truss  suitable 
for  spans  of  from  60  to  100  ft.  and  upwards. 

Arched  roofs,  Figs.  58  and  59,  admit  of  a  great  variety  of 
treatrrent.  They  have  a  pleasing  appearance,  and  cover  wide 
spans  without  intermediate  supports.  The  flatness  of  the 
arch  is  limited  by  the  requirement  of  a  minimum  thrust  at  the 
abutments.  The  thrust  may  be  resisted  either  by  thickening 
the  abutments  or  by  introducing  a  tie.  If  the  only  load  upon 
a  roof-truss  were  its  own  weight,  an  arch  in  the  form  of  an 
inverted  catenary,  with  a  shallow  rib,  might'  be  used.  But  the 
action  of  the  wind  induces  oblique  and  transverse  stresses,  so 
that  a  considerable  depth  of  rib  is  generally  needed.  If  the 
depth  exceed  12  in.,  it  is  better  to  connect  the  two  flanges  by 
braces  than  by  a  solid  web.  Roofs  of  wide  span  are  occasion- 
ally carried  by  ordinary  lattice-girders. 

Principals,  Purlins,  etc. — The  principal  rafters  m  Figs.  51 
to  57  are  straight,  abut  against  each  other  at  the  peik,  and  are 
prevented  by  tie-rods  from  spreading  at  the  heels.  When 
made  of  iron,  tee  (T),  rail,  and  channel  (both  single  1 — 1  and 
double  ][ )  bars,  bulb-tee  (T)  and  rolled  (I)  iron  beams,  are  all 
excellent  forms. 

Timber  rafters  are  rectangular  in  section,  and  for  the  sake 
of  economy  and  appearance,  are  often  made  to  taper  uniformly 
from  heel  t()  peak. 

The  heel  is  fitted  into  a  suitable  cast-iron  skew-back,  or  is 
fixed  between  wrought-iron  angle-brackets  (Figs.  60,  61,  62), 
and  rests  either  directly  upon  the  wall  or  upon  a  wall-plate. 


Fig.  61. 


Fig.  62. 


When  the  span  exceeds  60  ft.,  allowance  should  be  made 
for  alterations  of  length  due  to  changes  of  temperature.     This 


ROOF    TRUSSES. 


35 


may  be  effected  by  interposing  a  set  of  rollers  between  the 
skew-back  and  wall-plate  at  one  heel,  or  by  fixing  one  heel  to 
the  wall  and  allowing  the  opposite  skew-back  to  slide  freely 
over  a  wall-plate. 

The  junction  at  the  peak  is  made  by  means  of  a  casting  or 
w  rought-iron  plates  (Figs.  63,  64,  65). 


Fig.  63.  Fig.  64.  Fig.  65. 

Light  iron  and  timber  beams  as  well  as  angle-irons  are  em- 
ployed as  purlins.  They  are  fixed'  to  tire  top  or  sides  of  the 
rafters  by  brackets,  or  lie  between  them  in  cast-iron  shoes 
(Figs.  66  to  71),  and  are  usually  held  in  place  by  rows  of  tie- 


FlG.  70. 


Fig.  71. 


Fig.  72. 


rods,  spaced  at  6  or  8  ft.  intervals  between  peak   and  heel, 
running  the  whole  length  of  the  roof. 

The  sheathing  boards  and  final  metal  or  slate  covering  are 
fastened  upon  the  purlins.  The  nature  of  the  coverin'^^  regu- 
lates the  spacing  of  the  purlins,  and  the  size  of  the  purlins  is 
governed  by  the  distance  between  the  main  rafters,  which  may 


",i 


m 


36 


THEORY  OF  STRUCTURES. 


vary  from  4  ft.  to  upwards  of  25  ft.  But  when  the  interval 
between  the  rafters  is  so  great  as  to  cause  an  undue  deflection 
of  the  purlins,  the  latter  should  be  trussed.  Each  purlin  may 
be  trussed,  or  a  light  beam  may  be  placed  midway  between  the 
main  rafters  so  as  to  form  a  supplementary  rafter,  and  trussed 
as  in  Fig.  72. 

Struts  are  made  of  timber  or  iron.  Timber  struts  are 
rectangular  in  section.  Wrought-iron  struts  may  consist  of 
L-irons,  T-bars,  or  light  columns,  while  cast-iron  may  be  em- 
ployed for  work  of  a  more  ornamental  character.  The  strut- 
heads  are  attached  to  the  rafters  by  means  of  cast  caps, 
wrought-iron  straps,  brackets,  etc.  (Figs,  "Jt^  to  "jQ),  and  the 
strut-feet  are  easily  designed  both  for  pin  and  screw  connec- 
tions (Figs.  Ty  to  80). 


Fio.  7J. 


Fig.  74- 


Fig.  7s. 


Fig.  76. 


Fig.  77. 


Fig.  78. 


Fig   79. 


Fig.  80. 


1;     .rl! 


Tics  may  be  of  flat  or  round  bars  attached  cither  by  eyes 
and  pins  or  by  screw  ends,  and  occasionally  by  rivets.  The 
greatest  care  is  necessary  in  properly  proportioning  the  dimen- 
sions of  the  eyes  and  pins  to  the  stresses  that  come  upon  them. 

To  obtain  greater  security,  each  of  the  end  panels  of  a  roof 
may  be  provided  with  lateral  braces,  and  wind-ties  are  often 
made  to  run  the  whole  length  of  the  structure  through  the 
feet  of  the  main  struts. 


tn 


ROOF   WEIGHTS. 


37 


Due  allowance  must  be  made  in  all  cases  for  changes  of 
temperature. 

21.  Roof-weights. — In  calculating  the  stresses  in  the 
different  members  of  a  roof-truss  two  kinds  oi  load  have  to  be 
dealt  with,  the  one  permanent  and  the  other  accidental.  The 
permanent  "load  consists  of  the  covering,  the  framing,  and  ac- 
cumulations of  snozv. 

Tables  at  the  end  of  the  chapter  show  the  weights  of  various 
coverings  and  framings. 

The  weight  of  freshly  fallen  snow  may  vary  from  5  to  20 
lbs.  per  cubic  foot.  English  and  European  engineers  consider 
an  allowance  of  6  lbs.  per  square  foot  sufficient  for  sno'.v, 
but  in  cold  climates,  similar  to  that  of  North  America,  it  is 
probably  unsafe  to  estimate  this  weight  at  less  than  12  lbs. 
per  square  foot. 

The  accidental  or  live  load  upon  a  roof  is  the  wind-pressure, 
the  maximum  force  of  which  has  been  estimated  to  vary  from 
40  to  50  lbs.  per  square  foot  of  surface  perpendicular  to  the 
direction  of  blozc.  Ordinary  gales  blow  with  a  force  of  from  20 
to  25  lbs.,  which  may  sometimes  rise  to  34  or  35  lbs.,  and  even 
to  upwards  of  50  lbs.  during  storms  of  great  severity.  Press- 
ures much  greater  than  50  lbs.  have  been  recorded,  but  they 
are  wholly  untrustworthy.  Up  to  the  present  time,  indeed,  all 
wind-pressure  data  are  most  unreliable,  and  to  this  fact  may  be 
attributed  the  frequent  wide  divergence  of  opinion  as  to  the 
necessary  wind  allowance  in  any  particular  case.  The  great 
differences  that  exist  in  all  recorded  wind-pressures  are  pri- 
marily due  to  the  unphilosophic,  unscientific,  and  unpractical 
character  of  the  anemometers  which  give  no  correct  informa- 
tion either  as  to  pressure  or  velocity.  The  inertia  of  the  mov- 
ing parts,  the  transformation  of  velocities  into  pressures,  and 
the  injudicious  placing  of  the  anemometer,  which  renders  it 
subject  to  local  currents,  all  tend  to  vitiate  the  results. 

It  would  be  practically  absurd  to  base  calculations  upon 
the  violence  of  a  wind-gust,  a  tornado,  or  other  similar  phe- 
nomena, as  it  is  almost  absolutely  certain  that  a  structure 
would  not  lie  within  its  range.  In  fact,  it  maj'  be  assumed 
that  i'i  wind-pressure  of  40  lbs.  per  square  foot  upon  a  surface 


v'^B 


I 


i  III 


38 


THEORY  OF  STRUCTURES. 


perpendicular  to  the  direction  of  blow  is  an  ample  and  perfectly 
safe  allowance,  especially  when  it  is  remembered  that  a  greater 
pressure  than  this  would  cause  the  overthrow  of  nearly  all  the 
existing  towers,  chimneys,  etc. 

22.  Wind-pressure  upon  Inclined  Surfaces. — The  press- 
ure upon  an  inclined  surface  may  be  obtained  from  the  follow- 
ing formula,  which  was  experimentally  deduced  by  Hutton, 
viz, : 


M 


p„  =  p  sin  a'.84cosa-,. 


(A> 


p  being  the  intensity  of  the  wind-pressure  in  pounds  per  square 
foot  upon  a  surface  perpendicular  to  the  direction  of  blow,  and 
/>„  being  the  normal  intensity  upon  a  surface  inclined  at  an 
angle  a  to  the  direction  of  blow. 

Let  />/, ,  /j,  be  the  components  of  p„ ,  parallel  and  perpen- 
dicular, respectively,  to  the  direction  of  blow. 


Ph'=  Pn  sin  «,     and    p^  =  /„  cos  a. 


W 


Hence,  if  the  inclined  surface  is  a  roof,  and  if  the  wind  blows 
horizontally,  a  is  the  roof's  pitch. 

Again,  let  v  be  the  velocity  of  a  fluid  current  in  feet  per 
second,  and  be  that  due  to  a  head  of  Ji  feet. 

Let  w  be  the  weight  of  the  fluid  in  pounds  per  cubic  foot. 

Let  p  be  the  pressure  of  the  current  in  pounds  per  square 
foot  upon  a  surface  perpendicular  to  its  direction. 

If  the  fluid,  after  striking  the  surface,  is  free  to  escape  at 
right  angles  to  its  original  direction, 


V 


p  =  2]nv  =  — zv. 

Hence  for  ordinary  atmospheric  air,  since  w  =  .08  lb.,  approx- 
imately, 


p  =z    f'    = 

32  \,  20, 


(B) 


■     I 

III  I 


DISTRIBUTION  OF  LOADS. 


39 


When  the  wind  impinges   upon   a  surface  oblique  to  its 

.    Iv    sin  (i\ 
direction,  the  intensity  of  the  pressure  is  1 — -^ )  ,  v  being 

the  absolute  impinging  velocity,  and  yS  being  the  angle  between 
the  direction  of  blow  and  the  surface  impinged  upon.  (Sec 
chapter  on  Bridges.) 

Taoles  prepared  from  formulae  A  and  B  are  given  at  the 
end  of  the  chapter. 

23.  Distribution  of  Loads. — Engineers  have  been  accus- 
tomed to  assume  that  the  accidental  load  is  uniformly  dis- 
tributed over  the  whole  of  the  roof,  and  that  it  varies  from  30 
to  35  lbs.  per  square  foot  of  covered  surface  for  short  spans, 
and  from  35  to  40  lbs.  for  spans  of  more  than  6.")  ft.  But  the 
wind  may  blow  on  one  side  only,  and  although  its  direction  is 
usually  horizontal,  it  may  occasionally  be  inclined  at  a  con- 
siderable angle,  and  be  even  normal  to  a  roof  of  high  pitch. 
It  is  therefore  evident  that  the  horizontal  component  (/»/,)  of 
the  normal  pressure  (/„)  should  not  be  neglected,  and  it  may 
cause  a  complete  reversal  of  stress  in  members  of  the  truss, 
especially  if  it  is  of  the  arched  or  braced  type. 

If  P„  ^s  the  total  normal  wind-pressure  on  the  side  of  a 
roof  of  pitch  a,  its  horizontal  component  P^  sin  a  will  tend  to 
push  the  roof  horizontally  over  its  supports.  This  tendency 
must  be  resisted  by  the  reactions  at  the  supports. 

In  roofs  of  small  .span,  the  foot  of  each  rafter  is  usually^avrt^ 

to  its  support,  and  it  may  be  assumed  that  each  support  exerts 

P  sin  (y 
the  same  reaction,  which  should  therefore  be  equal  to  —" — . 

2 

In  roofs  of  large  span  the  foot  of  one  rafter  is  fi.vcd,  while  that 
of  the  other  rests  upon  rollers.  The  latter  is  not  suited  to  with- 
stand a  horizontal  force,  and  the  whole  of  the  horizontal  com- 
ponent of  the  wind-pressure  must  be  borne  at  the  fixed  end, 
where  the  reaction  should  be  assumed  to  be  equal  to  P„  sin  ci. 

In  designing  a  roof-truss  it  is  assumed  that  the  wind  blows 
on  one  side  only,  and  that  the  total  load  is  concentrated  at  the 
joints  (or  points  of  support)  of  the  principal  rafters. 

E.g.,  let  the  rafters  AB,  AC oi  a  truss  be  each  supported  at 


11 


11 


f 


1 

■ ;j-   I  : 

;<(!  .  It 


«-v...  .. '  'If 


m 


40 


THEORY  OF  STRUCTURES. 


two  intermediate  points  (or  joints),  D,  E  and  F,  G,  respectively, 
and  let  the  wind  blow  on  the  side  AB. 


Fig.  8i. 


;i    1' 


Take  BD  =  CF=l,,  DE  =  FG  =  l^,  EA  =  GA-l^\  and 
let  /, -f-^'j  + /a  = /;  .".  BC  —  2/ cos  or,  a  being  the  angle 
ABC. 

Let  W  be  the  permanent  (or  dead)  load  per  square  foot  of 
roof-surface. 

Let  pn  be  the  normal  wind-pressure  per  square  foot  of  roof- 
surface. 

Let  d  be  the  horizontal  distance  in  feet  from  centre  to 
centre  of  trusses. 

The  total  normal  live  load  concentrated 


at^^A'^^; 


at  Z>  =  p^d 


l.-Vh. 


at  E  —  p^d 


l.  +  l. 


atA=p„dj. 


\     The  total  vertical  dead  load  concentrated  at  D  and  F  = 


j^,^i±/? ;  at  £  and  C;  =  wd^^^'- ;  at  ^  =  wdl,, 

2  2  • 


Let  /?, ,  R^  be  the  resultant  vertical  reactions  at  B  and  C, 
respectively  (i.e.,  the   total  vertical   reactions   less   the  dead 

eights  \ivd-\  concentrated  at  these  points). 


w 


DISTRIBUTION  OF  LOADS. 


41 


Take  moments  about  C. 

:.  R^2l  cos  a  =  sum  of  moments  of  live  loads  about  C-\-  sum 
of  moments  of  dead  loads  about  C, 
^  moment  of  resultant  wind-pressure  about  C 
-f-  moment  of  resultant  dead  load  about  C, 

=  pjdi-  +  /  cos  2a j  +  wd{l^  -\-  2/,  -|-  2/,)/  cos  or, 


-where  — f-/cos  2a  is  the  perpendicular  from  C  upon  the  line 

of   action  of   the   resultant  wind-pressure   which   bisects  AB 
normally. 

(N.B.  The  moment  of  the  horizontal  reaction  at  ^  or  C 
about  C  is  evidently  nil.) 

R^  may  be  found  by  taking  moments  about  B, 

To  determine  the  stresses  in  the  various  members  of  a  roof- 
truss  two  methods  may  be  pursued  : 

{x)  A  single  stress  diagram  may  be  drawn  to  represent  the 
combined  effect  of  the  live  and  dead  loads.  This  will  be  found 
to  be  the  quickest  and  most  useful  method. 

{y)  The  normal  wind-pressure  (/>„)  may  be  resolved  into  its 
vertical  {p^)  and  horizontal  {p^  components ;  p^  may  then  be 
combined  with  the  dead  load  W,  and  a  stress  diagram  drawn 
for  the  vertical  loads  only.  A  second  diagram  may  be  drawn 
for  the  horizontal  loads.  The  resultant  stresses  will  be  the 
algebraic  sum  of  the  corresponding  stresses  in  the  two  dia- 
grams. 

A  third  method  will  be  referred  to  in  a  subsequent  article. 

24.  Ex.  I.  Method  {x)  applied  to  the  roof-truss  ^IBC, 
Fig.  82. 

The  dead  load  =  w/(^  concentrated  at  A. 

Id 
The  live  loads  =  p,~-  acting  at  each  of  the  points  A  and 

B,  normally  to  AB. 

The  vertical  reaction  at  B 


rvld        pjd  1 1       cos  2a 
'  2         cos  «\4   '        2 


)• 


M| 


li  i 


ii 


42 


THEORY  OF  STRUCTURES. 


Let  rollers  be  placed  underneath  C. 

The  total  horizontal  reaction  =  pjd  sin  a,  and  is  wholly 
borne  at  B» 


Fig.  83. 


■Si 


bciiu 


At 


given. 


Fig.  83. 

At  B  there  arejivc  forces  in  equilibrium,  of  which  three  are 
known,  and  the  reciprocal  of  B  may  be  thus  described  : 

Draw  5,5,  to  represent  the   normal  wind-pressure  (Ai"^) 

at  B ;  S,S^  to  represent  R,  ;  S,S^  to  represent  the  horizontal 
reaction  {p„M  sin  a) ;  S^S^  parallel  to  BD ;  ^'j^^  parallel  to 
AB. 

The  closed  figure  S^S^S^S^S^S^  is  the  reciprocal  required,  and 
the  stresses  in  BD,  AB,  at  B,  are  represented  by  5,5^,  S^S^, 
respectively,  being  a  tension  and  a  thrust. 

At  D  there  are  three  forces  in  equilibrium,  of  which  the 
tension  in  DB  has  been  found.  Drawing  5,5„  horizontally 
and  5j5,  parallel  to  AD,  the  triangle  S^S,S^  is  evidently  the 
reciprocal  of  D,  the  stresses  in  DA,  BE  being  represented  by 
5j5g,  6",6\,  respectively,  and  being  both  tensions.  , 


ROOF- TRUSSES. 


45 


by 


Again,  the  triangle  S^S^S^  is  the  reciprocal  of  E,  the  stressc  s 
in  EC\  EA  being  represented  by  5,5, ,  5,5, ,  respectively,  and 
being  both  tensions. 

At  A  there  are  six  forces  in  equilibrium,  of  which  two,  viz., 

/    ld\ 
the  normal  pressure,  [pn"^],  and  the  dead  weight,  {wld),  arc 

given,  while  the  stresses  in  AB,  AD,  AE  have  been  found. 

Id 
Draw  S^S^  to  represent  /„ —  ,  and  S^S^  to  represent  wld. 

Five  of  the  forces  at  A  are  therefore  represented  by  the 
following  lines,  taken  in  order  :  S,S.,,  S^S, ,  S,S^,  S,S,,  S^S, . 

Hence  the  closing  line  5,5^  must  necessarily  represent  in 
direction  and  magnitude  the  force  in  AC  at  A,  and  it  is  a 
thrust. 

Also,  5,5^5,  must  be  the  reciprocal  of  C,  and  therefore  S^S^ 
represents  the  reaction  at  C. 

T/ie  resultant  reaction  at  B  is  represented  in  direction  and 
magnitude  by  S^S^ . 

The  line  SJS^  must  pass  through  the  point  S^ ,  as  535, ,  the 
horizontal  reaction,  is  merely  the  horizontal  projection  of  S^S^ , 
the  total  wind-pressure. 

The  dotted  lines  show  the  altered  stresses  if  rollers  are 
under  /j,  the  end  C  being  fixed.  The  stress  in  each  member  is 
diminished,  and  as  the  truss  should  be  designed  to  meet  tlie 
most  unfavorable  case,  the  stresses  should  be  calculated  on  the 
assumption  that  the  rollers  are  on  the  leeward  side. 

This  may  be  considered  an  invariable  rule  for  roof-trusses. 

Ex.  2.  Method  (x)  applied  to  the  roof-truss  ABC,  Fig.  84. 

The    vertical    uead    load  = at    each    of    the    pomts 

E,  A,  G. 


The  live  load,  acting  normally  to  AB,  = 


pjd 


at  each  of 


the  points  B  and  A,  and 


pJd 


at  F. 


The  vertical  reaction  R^  at  B 


=  \zvld  -{-  :77^—  (cos  2»  4"  i)* 


2  cos  a 


liKjfF*^" 


r '1- 


>1.  v\\ 


Iti 


44 


THEORY  OF  STRUCTURES. 


I     1 


The  horizontal  reaction  at  ^  =  pjd  sin  a,  rollers  being 
under  C  as  before. 


Fig.  85. 


Describe  the  stress  diagram  in  precisely  the  same  manner 
as  in  Ex.  i. 

Taking  S^S^  to  represent  the  normal  wind-pressure  at  jff, 
5,5,  " 


"    vertical  reaction  R^  at  B, 
"    horizontal  reaction  at  B, 


.-.  S,S^S,S,S,S,       is  the 

reciprocal  of  B, 

SAS.S,S,S, 

"  F, 

S,S,S,S,S, 

"  D, 

0,0,0,0,„0,,0,aO, 

"  ^, 

5,,>j,o>Ji3'Ju5,, 

"  G^, 

'^9'^*'^18'^10«^» 

"  j&, 

545,40,3 

«  C. 

5,5^  is  the  horizontal  projection  of  S^St  +  SiS,  -\-  S^S^^ ,  i.e., 
of  the  total  normal  wind-pressure,  and  therefore  the  vertical 
through  5„  must  pass  through  5« . 


J1:: 


ROOF   TRUSSES. 


45 


The  dotted  lines  show  the  altered  stresses  if   rollers  are 

under  B. 

The  resultant  reaction  at  B  is  represented  in  direction  and 

nia;4iiitLide  by  S^S^. 

Ex.  3.  Method  {x)  applied  to  the   truss    represented   by 

Fig.  86. 


Fig.  86. 

Data.  —Pitch  =  30° ;  AD  -  BD  -  AE  -  CE  =  21  ft. ; 
trusses  13  ft.,  centre  to  centre  ;  dead  weight  =  8  lbs.  per  square 
foot  of  roof-surface  ;  wind-pressure  on  one  side  of  roof  (say  AE) 
normal  to  roof-surface  =  28  lbs.  per  square  foot;  DF=DH 
=  EG  —  EK\  DF  and  EG  are  vertical  ;  rollers  under  one  end, 
say  C\  span  =:  79  ft. ;  AF=.BH  =  21  ft.,  nearly  ;  FN  =  3^  ft, 
nearly. 


Total   live   load 


and  = 


Total  dead  load  = 


4459  lbs.  f= 13 -28]  at    each    of   the 

points  F,  H, 

3822  lbs.(  =  —  .13.28]    at    each    of    the 
points  A,  B. 

1274  lbs.(=:-^*.  13.8)   at    each    of    the 
points  F,  //,  K,  G, 


and  =  2184  lbs.  (  =  21  .  13  .  8)  at  the  point  A. 


I 


:. 

I  lit 


.■  '1 


.1.    ' 


:'i 


.  \ 


m 


i  ii; 


46 


THEORY  OF  STRUCTURES. 


«Jl 


t/3/ 


i    III     -:.M 
«l    if 


CO 


S   2   2;     ^ 


CO 


OT 


ot«"ot  wT     ^ 


ROOF   TRUSSES.  47 

Resultant  vertical  reaction  at  B 

=  K4  X  1274  +  2184)  +  ^^-  =  13201.8  lbs. 

Horizontal  reaction  at  B  =  16562  sin  30°  =  8281  lbs. 

Let  I  inch  represent  16,000  lbs.,  and  on  this  scale  draw 

5,5j  =  3822  lbs.,  the  normal  wind-pressure  at  B  ; 
S^S^  —  13201.8  lbs.,  the  vertical  reaction  at  B ; 
S.,S,  =  8281  lbs.,  the  horizontal  reaction  at  B  ; 
StS^  parallel  to  BD,    and    s^s^  parallel  to  BA. 

The  figure  S^S^S,S^S^S^  is  the  reciprocal  of  B. 
The  stress  diagram  can  now  be  easily  completed,  the  recip- 
rocals of  the  points  H,  F,  D,  A,  G,  K,  E,  and  C  being 

5,e5„5„5.,5.„ ,    S,,S,^S,,S,^S,,,    and    S,^S,^S,S,,,    respectively. 

St ,  as  before,  is  in  the  vertical  line  S^^S^^  produced. 
On  the  assumed  scale, 


'■'dm 


,  t 
i  \ 

t 

tfil 


5«5j  =  the  tension  at  BD ; 

S,A,=    "        "        "  AD; 

5,5.,=   "        "        "  BE; 


s,s,= 

thrust 

inBH 

SA  = 

ki 

"  HF 

5g>Jii    = 

(< 

"   AF 

s.s,= 

<< 

"  DH 

s,s,= 

(( 

"  DF. 

These  are  the  maximum  stresses  to  which  the  members  of 
me  half  of  the  truss  can  be  subjected,  and  for  which  they 
should  be  designed.  It  is  also  usual,  except  in  special  cases,  to 
make  the  two  halves  symmetrical. 

5,5,  is  the  resultant  reaction  at  B. 

If  the  end  C  is  fixed  and  rollers  placed  under  B,  the  reduced 
stresses  may  be  shown  by  dotted  lines  as  in  Exs.  i  and  2. 


■if! 


"-^M 


48 


THEORY  OF  STRUCTURES. 


Exs.  4  and  5.  Method  {x)  applied  to  the  trusses  repre- 
sented by  Figs.  88  and  90.  • 

It  is  assumed,  as  before,  that  there  is  a  normal  wind-press- 
ure upon  ABy  and  that  rollers  are  under  C. 

Figs.  89  and  91  are  the  maximum  stress  diagrams  corre- 
sponding to  Figs.  88  and  90,  respectively,  and  are  drawn  in  pre- 
cisely the  i-ame  manner  as  described  in  the  preceding  examples. 

Remark  on  Fig.  88. — The  stresses  at  the  joints  F  and  D 
are  indeterminate,  and  it   is  assumed  that  the  stress  in  FL 


Fig.  88. 


Fig.  89. 

is  equal  to  that  in  FH.     The  reciprocal  of  F  thus  becon 
6",„5„5e5g5,5„5,35"„,  5„5,3(=  S,S,)  being  the  stress  in  FD.     Tl 
truss  is  an  example  of  a  frame  with  redundant  bars,  in  which 
the  stresses  can  only  be  determined  when  the  relative  yield  of 
the  bars  is  known. 


ii 


im 


ROOF   TRUSSES. 


49 


Remark  on  Fig.  90. — The  stress-diagram,  Fig.  91,  for  each  of 
the  joints  in  the  horizontal  BC  (Fig.  90)  is  closed  by  the  return 
of  one  side  upon  another.  Thus  at  D  the  stress  diagram  is 
S,S,StS^S, ,  the  closing  line  5,6",  (the  tension  in  D£)  returning 


3^ 

f^ 

u 

[X, 

c 

^^^                      D         E 

'\1 

^ 

^ 

^ 

^- 

Fig.  gi. 

upon   5^5",  (the   tension  in  DB).    The  total  stress  in  AE  is 
evidently  represented   by  S^^S^^,  the   reciprocal   of  A    being 

Ex.  6.  A  truss  with  curved  upper  and  lower  chords,  the 
portions,  however,  between  consecutive  joints  being  assumed 
straight. 

Under  a  uniformly  distributed  load  the  truss  (Fig.  92)  is 
evidently  incomplete,  and  the  stress  diagrams  at  the  joints  in 
the  lower  chord  will  not  close,  so  that  equilibrium  is  impossible. 
T'l''  frame  is  made  complete  and  the  stresses  determinate  by 
inuouucing  ties  as  in  Fig.  93,  the  corresponding  stress  diagram 
for  OP"  half  the  truss  being  shown  by  Fig.  94. 

N  t,  let  there  be  a  wind-pressure  on  the  side  AB  of  the 
truss.  In  order  to  prevent  a  reversal  of  stress  in  the  diagonal 
ties  on  the  side  AC  (Fig.  93),  additional  ties  DE,  FG,  called 
coimter-braus,  arc  introduced  as  in  Fig.  95.     Fig.  96  gives  the 


.!»! 


1 


■m 


.,.-,.  a 


giSfSM 


L       'i 


50 


THEORY  OF  STRUCTURES. 


stress  diagram  due  to  wind-pressure  only,  it  being  assumed 
that  the  end  C  rests  upon  rollers  and  that  B  is  fixed. 


Pic.  93. 


Fig.  93. 


Fig.  94. 


Fig.  95. 


Fig.  q6. 


Note. — 21  =  wind-press,  at  .5  =  f  wind-press,  upon  BM, 
23  = 

34  = 

45  = 
56  = 

67  = 


"M  = 

'^31=-. 

^'  0  = 

"  6>  = 

.    "  A  = 

BM, 
MO, 
MO, 
OA, 
OA. 


m 


\K  =  vertical  reaction  at  B, 
HK  =  horizontal  reaction  at  B. 

Ex.  7.  A  single  example  will  serve  to  illustrate  method  (7). 
Take  the  truss  represented  by  Fig.  97. 

Fig.  98  is  the  stress  diagram  due  to  the  vertical  load  upon 
the  roof,  viz.,  the  dead  weight  -\-  vertical  component  of  wind- 
pressure,    />g  is  the  vertical  reaction  at  B  and  is 


4  4 


ROOF-TRUSSES, 
qni  is  the  weight  at  F  and  is 

p„   cos  a-\-  w 


BH  =  weight  at  //"  =  inn. 


51 


Fig.  97. 


Fig.  98. 


Fig.  99  is  the  stress-diagram  due  to  horizontal  component 
of  wind-pressure,  rollers  being  placed  under  B  and  the  end  C 
beincr  fixed. 


p'o  =  downward  reaction  at  B 


pjd  sin"  n 


4    cos 


tx 


,  ,       ,      .          ,  ,           ,     .    ,        ,,      A,</ sm  ot 
0  q  —  honzontul  force  of  wmd  at  B  — BF\ 

q'm  =  m'n'=  horizontalforceof  windat  For//=:  —^ BH. 


I  nil 


m 


liUt. 


niy 


Total  resultant  stresses  in  the  members  Bl .  Fil,  HA,  DF^ 
DH,  BB,  DA,  DE  are  represented  by  qr  —  q'r',  ms  —  m' s\ 
lit  —  n't',  sr  —  s'r',  st  —  s't',  pr  —  p'r',  tv  —  t'v',  pv  —  p'v\ 
respectively. 

Note. — The  stress  diagrams  for  trusses  with  both  of  the  lower 
<.ik1s  of  the  principal  rafters  fi.val,  are  drawn  in  precisely  the 
same  manner  as  described  in  the  preceding  examples. 


!K''i  I 


^ 


-4- 


H 


52 


THEORY  OF  STRUCTURES. 


i         ■• 


m  li 


:f  -I 


Thus,  in  Fig.  100,  S.S^S^S^S^S^  is  the  reciprocal  of  A,  S^S, 
representing  the  portion  of  the  horizontal  wind-pressure  borne 


Fig.  too. 

at  A.  Again,  HSJS^S^H  is  the  reciprocal  of  B,  NS,  represent- 
ing the  portion  of  the  horizontal  wind-pressure  borne  at  C. 
HS,  =  HS^  -\~  S^S^  =  total  horizontal  wind-pressure,  S^S^  repre- 
senting the  vertical  reaction  at  B,  and  HS,  that  at  C. 

25.  Bridge-trusses. — A  bridge-truss  proper  consists  of 
an  upper  cliord  {ox  flange),  a  lower  chord  (or  flange),  and  an  in- 
termediate portion,  called  the  zveb,  connecting  the  two  chords. 
Its  depth  is  made  as  small  as  possible  consistent  with  economy, 
strength,  and  stiffness.  Its  purpose  is  to  carry  a  distributed 
load,  which,  as  in  the  case  of  roof-trusses,  is  assumed  to  be 
concentrated  at  the  joints,  or  panel-points,  of  the  upper  and 
lower  chord.  Trussed  beams  are  also  employed  for  the  same 
object,  and  examples  of  simple  frames  of  this  class  have  already 
been  given. 

The  following  are  bridge-trusses  of  a  more  complex  char- 
acter. 

Ex.  I.  The  beam  i9C(Fig.  loi)  is  supported  at  three  points 
by  the  vertical  struts  DF,  AK,  EG,  which  are  tied  at  the  feet 
by  the  rods  DB,  DK,  AB,AC,  and  EK,  EC.  Let  W„  IV„  W, 
be  the  loads  concentrated  at  the  joints  F,  K,  G,  respectively. 
Draw  the  line  of  loads  S^S^ ,  S^S^  being  W^ ,  S,S^  =  W^ ,  and 

Describe  the  funicular  polygon  with  any  pole  0,  and  draw 
OH  parallel  to  the  closing  line  MN  of  this  polygon.     Then 


BRIDGE    TRUSSES. 


53 


HS^  is  the  reaction  at  B  and  HS^  the  reaction  at  C  (Art.  3). 
HS^S^  is  the  reciprocal  of  B,  S^S^  being  the  thrust  along 

FB,  and  S^H  the  tension  along  BD. 
S^SjSjS^S^  is  the  reciprocal  of  F,  S^S,  being   W^,  the 
weight  at  F,  5,5,  the  thrust  along  KF,  5,5,  the 
thrust  along  DF. 
HS^S^S,/!  is  the  reciprocal  of  D,  S^S^  being  the  tension 

along  DK,  and  5,/f  the  tension  along  Z)^. 
HS^S^H  is  the  reciprocal  of  A,  S^S^  being  the  thrust  along 
KA,  and  58//"  the  tension  along  AE. 
So,  5,5,5,5e5,5,5, ,  S^S^S^^S^S^ ,  S^S.^HS^S^ ,  and  S^HS,^  are 
the  reciprocals  of  K,  G,  £,  and  C,  respectively,  the  closing  line 
5,„54  being  necessarily  horizontal  and  representing  the  stress 
in  GC. 


-z^O 


Fig.  ioi.  Fio.  loa, 

This  truss  inverted  is  often  used  for  bridge  purposes  in  dis- 
tricts where  timber  is  plentiful,  as  it  .  y  be  constructed 
entirely  of  wood.  The  stresses  in  the  sevetal  members  of  the 
inverted  truss  are  of  course  reversed  in  kind  but  unchanged  in 
magnitude,  and  are  given  by  the  same  stress  diagram. 

Note. — The  reactions  //5, ,  HS^  may  be  obtained  at  once  by 
the  method  of  moments.  Thus,  by  taking  moments  about  C, 
the  reaction  R^  at  B  is 

and  by  taking  moments  about  B,  the  reaction  R,  at  C  is 


!■  .:i 


\l  i 


m 


£i 


I  i 


11 


54 


THEORY  OP   STRUCTURES. 


Ex.  2.  In  the  truss  represented  in  the  accompanying  figure, 
the  length  of  the  beam  AB  is  so  great  that  the  single  triangu- 
lar truss  ACB  with  a  single  central  strut  CO  is  an  insufficient 
support.  The  two  halves  are  therefore  strengthened  by  tlie 
simpie  triangular  trusses  AGO  with  a  central  strut  GF  and 
BPO  with  a  central  strut  PN. 

Again,  each  quarter-length,  viz.,  AF,  FO,  ON,  NB,  is  simi- 
larly trussed.  The  subdivisions  may,  if  necessary,  be  carried 
still  farther.     This  truss  in  four,  eight,  dxteen,  .  .  .  divisions  or 

D  F    S     H  0  I.  N  Q  B 


K  C  ,1/ 

Fig.  103. 

panels  is  known  as  the  Fink  truss,  and  has  been  v/idely  em- 
ployed in  America,  the  number  of  panels  usually  being  eight 
or  sixteen. 

The  members  shown  by  the  dotted  lines  may  be  introduced 
for  stiffness,  and  the  platform  may  be  either  at  the  top  or 
bottom.  The  weight  directly  borne  by  a  strut  is  usually  de- 
termined from  the  loads  upon  the  two  adjacent  panels  by 
assuming  the  corresponding  portions  of  the  beam  to  be  inde- 
pendent beams  supported  at  the  ends.  Thus  if  there  be  a 
weight  W-'at  the  point  .S  in  the  panel  FH,  the  portion  of  W 
borne  by  the  strut  GF  at  F  is 

SH 


W 


FH' 


and  the  portion  borne  by  the  strut  KH  at  //"  is 

/re  I   , 

^  FN' 

Let  W,,  W„  W,,  IV„  W,,  W„  IV,  be  the  weights  upon 
the  struts  (or  posts)  DFS,  FG,  HK,  OC,  LM,  NP,  QR,  respect- 
ively. 

Let  P,,  P,,  Pt,  P^,  P^,  P^,  P,  be  the  compressions  to  which 
these  posts  are  severally  subjected. 


T.= 


T.= 


Z  = 


\r 


:^1ii-' 


FINK   TRUSS. 


55 


Let  a,  /?,  y  be  the  inclinations  to  the  vertical  of  AE,  AG, 
AC,  respectively. 

Let  T^,  1\,  T^,  .  .  .  be  the  tensions  in  the  ties, as  in  Fig.  103. 

The  tensions  in  the  ties  meeting  at  the  foot  of  a  post  are 
evidently  equal. 

Each  triangular  truss  may  be  considered  separately. 

From  the  truss  ^ii/%  2  7",    cos  a  =  P,  =  W^\ 

from  the  truss  A  GO,  2  T^    cos  p=  P^=  ^^j+(  ^1+  ^«)  cos  a  ; 

from  the  truss  FKO,  2  T^  cos  a  =  P^—  W^; 
from  the  truss  A  CB, 

2T,co^y  =  P,=  ^f',-L(7;+  r,)cos^  +  (r,-f  r.)cosa; 

from  the  truss  OMN,  iT^cos  a  =  P,=  W^; 

from  the  truss  OPB,  2  T,  cos  /?  =P,=  W,-^{  r,+  T,)  cos  a  ; 

from  the  truss  NRB,  2T^Q0sa  =  P,  =  W^. 
Hence 

W 
r  =  — -'sec«, 
2 


,.  =  V,,+  -.d-_-.)3ec,. 


riH 


Hn 


lich 


T.=  —  sec  a, 


r. 


w;  + 


W^^rW.  ,    W^.  +  3f^3+3^.+  M^'0 


-^  + 


4 


sec  y, 


r  =---seco', 


r.  =  i(«'.+  «^±l'^)secA 


^      W',  ..... 

/,  = sec  or, 

and  the  values  of  P, ,/',,/',,..  .  can  be  at  once  found. 


56  THEORY  OF  STRUCTURES. 

Again,  the  thrust  along  AF-=  7",  sin  a-\-T^  sin /3  -\-  T^  sin  y  J 
"  ^X  F—  7", sin  fi-\-T^svs\Y', 

"  along  F0=  7",  sin  /S  -f-  /"^  sin  ;/  -|~  ^i  sin  a  j 

"  at  d?  =  T^  sin  ;/ ; 

etc.,        etc. 

If  the  truss  carries  a  uniformly  distributed  load  W, 
W,=  W,=  W,=  W,=  W,=  W,=  W,=  ^; 

w 

T,=  T^=T^=T,  =  -^  sec  a, 

W  W 

7;  =  7;  =  —  sec  /?,      ^«  =  "T  sec  ;/. 


1  1 


If  the  above  diagram  is  inverted,  it  will  represent  another 
type  of  truss  in  which  the  obliques  are  struts  and  the  verticals 
ties. 

Note. — The  stresses  in  the  several  members  of  each  of  the 
trusses  due  to  the  weight  it  is  designed  to  carry,  may  of  course 
be  easily  determined  graphically  in  the  manner  already  de- 
scribed in  previous  articles, 

Ex.  3.  Fig.  104  represents  a  beam  trussed  by  a  number 
of   independent  triangular  trusses,   the   vertical   posts    being 


F;g.  104. 

equidistant.  The  weight  concentrated  at  the  head  of  each 
post  may  be  found  by  the  method  described  in  Ex.  2,  which 
in  fact  is  generally  applicable  to  all  bridge  and  roof  trusses. 

Let  7,,  7,  be  the  tensions  in  AE,  BE,  respectively. 

Let  W^  be  the  weight  at  D. 


W^: 


WARREN   TRUSS. 


57 


Let  a, ,  a^  be  the  inclinations  of  AE,  BE,  respectively,  to 
the  vertical. 


r.=  w. 


sin  a. 


sin  (a,  +  or,) ' 


sinK  +  a,) 


Similarly,  the  stress  in  any  other  tie  may  be  obtained. 

The  compression  in  the  top  chord  is  the  algebraic  sum  of 
the  horizontal  components  of  all  the  stresses  in  the  ties  which 
meet  at  one  end. 

The  verticals  are  always  struts  and  the  obliques  ties. 

This  truss  has  been  used  for  bridges  of  considerable  span, 
but  the  ties  may  prove  inconveniently  long. 

Ex.  4.  The  figure  SANT  represents  an  ordinary  triangular 
truss  of  the  Warren  type,  supported  at  the  ends  5  and  T. 

Draw   the   line  of  loads    16,  12      a      c      e      g      l      n 
being   the  weight  at  B,  and  23,  34,  /\}\7\Jf\/\}\ 
45,  56  the  weights  at  A  F,  K,  M,  §    %'    'd'    ' f'    'k'    'm'    'V 
respectively. 

With  any  pole  O  describe  the 
funicular  polygon  and  draw  OP  par- 
allel to  its  closing  line  QR. 

.•.  iP  is  the  reaction  at  S,  and  6P 
that  at  T. 

The  reciprocal  of  5  is  the  triangle 
PiS,;  1 5,  being  the  tension  in  SB, 
and  5, /'the  compression  in  AS. 

The  reciprocal  of  A  is  the  triangle  Q 
PS,S, ;  5,5,  being  the  tension  in  AB, 
and  ^jP  the  compression  in  CA. 

The  reciprocal  of  B  is  the  figure 
S^iiS^SjS,;  25,  being  the  tension  in 
BD,  5,5,  the  compression  in  CB,  and 
12  the  weight  at  B.  fio.  107. 

The  reciprocal  of  C  is  the  figure  PS^S^S^P;  5,5,  being  the 
tension  in  CD,  and  S^P  the  compression  in  EC. 

The  reciprocal  of  D  is  the  figure  5,235j5«5, ;  35,  being  the 
tension  in  DF,  5^5,  the  compression  in  ED,  and  23  the  weight 
atZ). 


Fig.  T06. 


Km 


111 


,  <■'.  ,«i 


58 


THEORY  OF  STRUCTURES. 


% 


'! 


The  reciprocal  of  E  is  the  figure  PS^S^S^P;  5,5,  being  the 
tension  in  EF,  and  S^P  the  compression  in  GE. 

The  reciprocal  of  Fis  the  figure  S^^^^S.S^S^ ;  45,  being  the 
tension  in  FK,  5,5",  the  tension  in  FG,  and  34  the  weight  at  F. 
And  so  on,  the  closing  line  PS^,  for  the  reciprocal  of  T  being 
necessarily  parallel  to  NT. 

The  arrow-heads  show  the  character  of  the  stresses  in  the 
several  members  of  the  truss. 

No/e. — The  reactions  may  also  be  at  once  determined  by 
the  method  of  moments. 

Thus        iP  =  1(12)  +  1(23)  +  K34)  +  1(45)  +  i(56), 
and  6P  =  K12)  +  1(23)  +1(34)  +  t(45)  +  K56). 

Ex.  5.  In  the  truss  represented  by  the  accompanying  figure, 

the  joints  in  the  upper  as  well  as 
those  in  the  lower  chord  are  loaded, 
the  weights  being  transmitted  to  the 
former  by  means  of  vertical  sus- 
penders. 

Fig.  109  is  evidently  the  corre- 
sponding stress  diagram. 

JVoic.  —  In  the  trusses  repre- 
sented by  Figs.  106  and  109,  the  floor 
is  carried  upon  the  lower  chords.  If 
the  trusses  are  inverted,  the  floor 
may  be  carried  on  the  upper  chords. 
The  stresses  in  the  several  members 
are  evidently  the  same  in  magnitude 
and  are  only  reversed  in  kind. 
f"'°-  "o?-  Ex.  6.  The  Howe   truss  repre- 

sented by  Fig.  1 10  is  verj'  widely  used  and  may  be  constructed 
of  timber,  of  iron,  or  of  timber  and  iron  combined. 

■     A       C.      E,      G       L,      N.     ,R 


(lii)      (23)      m     (46)     (66)     (CJ)      (TS) 

Fig.  108. 


HOWE    TJiUSS. 


59 


Let  there  be  a  uniformly  distributed  load  upon  the  truss 
consisting  of  a  weight  W  at  each  of  the  joints  B,D,  .  .  .'in  the 
lower  chord. 

The  reaction  at  each  support  =  ^^W. 

Fig.  1 1 1  is  the  stress  diagram,  and  the  several  members  of 
the  truss  are  indicated  on  the  lines  representing  the  stresses  to 


BS 


■  ^ 

w 

/BD 

5 
^ 

1 

w 

./ 

> 

CD 

DF     / 

w/ 

<§/  8 

/    FH 

/ 

/          AC 

/ 

/      < 

r 

\               c^ 

\      EG 

\         - 

^ 

w\ 

<^\      HK 

km\ 

\ 

w 

\ 

MQ 

w 

QT 

\ 

Fig.  III. 

which  they  are  subjected.  The  directions  of  these  stresses  at 
the  joints,  and  hence  also  their  character,  are  easily  determined 
by  following  in  order  the  sides  of  the  reciprocals.  The  verti- 
cals are  evidently  all  ties  and  the  diagonals  all  stmts. 

If  the  load  is  unevenly  distributed,  the  stresses  in  different 
members  may  be  reversed.     For  example, 


Fig.  113. 


Let  the  truss  carry  a  single  weight  P  at  any  point  D. 

The  reciprocal  of  D  is  S^S.,S^S,S^S,  (Fig.  112),  5,5,  represent^ 


H>     t 


i  !    \\ 


1.    * 


«o 


THEORY  OF  STRUCTURES. 


;    ! 


ing  P,  and  the  arrow-heads  showing  the  directions  of  the  forces 
now  acting  at  D.  Thus  the  force  in  DE  at  Z>,  represented  by 
5,5^ ,  acts  from  D  towards  E,  and  is,  therefore,  a  tension. 

Hence,  in  order  that  DE  may  not  be  subjected  to  a  tensile 
force,  counterbraces  CF,  EH  are  introduced  so  that  the  por- 
tion of  P  borne  on  the  support  at  T  may  be  transmitted 
through  the  system  CFEH  to  H  and  from  H  to  T  through  the 
regular  system  HGKLMNQRT.  The  reciprocal  of  D  is  now 
,S,5,S,5,  (Fig.  1 1 3),  and  the  reciprocal  of  C  the  figure  HS^S^S^S^H, 
the  arrow-heads  showing  the  directions  of  the  forces  at  C.  It 
will  be  at  once  observed  that  FC  must  be  a  strut. 

In  order  to  make  provision  for  a  varying  load,  as  when  a 
train  passes  over  a  bridge,  counterbraces  are  introduced  in  the 
panels  on  both  sides  of  the  centre,  and  although  they  may  not 
be  necessary  in  every  panel,  they  will  give  increased  stiffness  to 
the  truss. 

Note. — Generally  speaking,  a  panel  is  that  portion  of  the 
bridge-truss  between  two  consecutive  verticals,  and  the  ends 
of  the  verticals  are  called  panel-points. 

Ex.  7.  Fig.  1 14  represents  a  Pratt  truss,  and  is  merely  an 
inverted    Howe  truss.     The  diagonals   become   ties  and  the 


t 

II' 


Pig.  1x4. 

verticals   struts.     Counterbraces   are  introduced  to  resist  the 
action  of  a  varying  load,  precisely  as  described  in  Ex.  6. 

Ex.  8.  The  bowstring  truss  in  its  simplest  form  is  repre- 


FlG.  lis. 


sented  by  Fig.  115.    Assuming  that  the  portions  of  the  upper 
chord  between  consecutive  joints  are  straight,  the  stress  dia- 


BOWSTRING  TRUSS. 


6i 


gram  for  a  uniformly  distributed  load  and  for  one  half  the 
truss  is  Fig.  Il6. 

The  panels,  however,  are  incomplete  frames,  and  if  the  truss 

Ss , S, 


Fig.  ii6. 


Fig.  117. 


has  to  carry  an  unequally  distributed  load,  ties  similar  to  that 
shown  by  the  dotted  line  MNviwxsX.  be  introduced  in  the  several 
panels  in  order  to  prevent  distortion. 

For  example,  let  there  be  a  single  load  P  at  the  joint  N, 
and  let  there  be  no  brace  NM.  The  stress  in  the  first  vertical 
is  evidently  nil.  The  reciprocal  of  iV  is  S^S^S^S^S^S^ ,  Fig.  117, 
S^S^  representing  P.  The  reciprocal  of  L  is  HSJS^S^H,  and  the 
arrow-heads  show  the  directions  of  the  forces  at  H. 

Thus  the  force  in  6^Z,  which  is  represented  by  S^S, ,  acts  from 
0  towards  Z,  and  is,  therefore,  a  compression.  But,  under  a 
uniformly  distributed  load,  the  diagonals  are  all  ties,  and  NM 
is  introduced  to  take  up  that  portion  of  P  which  would  be 
otherwise  transmitted  through  LO  in  the  form  of  a  compression. 
In  this  case  the  reciprocal  of  L  is  HS^S^H,  since  the  stress  in  LO 
due  to  P  is  assumed  to  be  nil.  Also  the  reciprocal  of  N  is 
S,S.,S^S^S^S^S^.  The  stress  in  NM,  represented  by  ^g^, ,  acts 
from  A^to  J/ and  is  a  tension. 

Hence  the  diagonals  NM  are  also  ties,  and  the  portion  of 
the  weight  P  borne  at  L  is  carried  to  Q  through  the  system 
NAIOQ. 

Ex.  9.  Fig.  118  is  a  bowstring  truss  with  isosceles  bracing. 
Under  an  arbitrary  load  Fig.  119  is  the  stress  diagram,  the 
loads  at  a,  b,  c,  d,  e,f,  g  being  12,  23,  34,  45,  56,  67,  78,  respect- 
ively. As  in  the  Warren  girder,  the  diagonals  may,  under  the 
action  of  a  varying  load,  be  subjected  to  both  tensile  and  com- 


% 


li 


f 

I   \ 


\:'\\ 


hi    i 


^2 


THEORY  OF  STRUCTURES. 


pressive  stresses.     They  mut.f,  therefore,  be  designed  to  bear 
such  reversal  of  stress. 


ah        c        a        e        J        (/ 
Fig.  ii8. 


p 


Fig.  119, 

It  is  assumed,  as  before,  that  the  portions  of  the  upper 
chord  between  consecutive  joints  are  straight. 

Note. — The  design  of  bridge-trusses  will  be  further  con- 
sidered in  a  subsequent  chapter. 

26.  Method  of  Sections. — It  often  happens  that  the 
stresses  in  the  members  of  a  frame  may  be  easily  obtained  by 
the  method  of  sections.  This  method  depends  upon  the 
following  principle : 

If  a  frame  is  divided  by  a  plane  section  into  two  parts,  and 
if  each  part  is  considered  separately,  the  stresses  in  the  bars 
(or  members)  intersected  by  the  secant  plane  must  balance  the 
external  forces  upon  the  part  in  question. 

Hence  the  algebraic  sums  of  the  horizontal  components, 
'2{X\  of  the  vertical  components,  "^{Y),  and  of  the  moments 
of  the  forces  with  respect  to  any  point,  2{M),  are  severally 
zero ;  i.e.,  analytically, 

2{X)  =  o,    2(y)  =  o,    and     2{M)  =  o. 

These  equations  are  solvable,  and  the  stresses  therefore 
determinate,  if  the  secant  plane  does  not  cut  more  than  three 
members. 


EXAMPLES,  63 

Ex.  I.  ABC  is  a  roof-truss  of  60  ft.  span  and  30°  pitch. 


Q       N 

4KTOtl8 

Fig.  120 

The  strut  DF  =  GH  =  i,  ft.;  the  angle  FDA  =90°.  Also 
AF=FB  =  AG=  GC. 

The  vertical  reaction  at  ^  =  5  tons.  The  weight  concen- 
trated at  Z>  =  4J-  tons. 

Let  the  angle /i^F=  a. 


AB  —  30  sec  30°  —  20  ^^3  ;     cot 


ioi/3 

a  = = 


2V3, 


sin  a  — 


V13 


cos  a  z= 


2i^ 
VY3' 


If  the  portion  of  the  truss  on  the  right  of  a  secant  plane 
A/N  be  removed,  the  forces  C,  7", ,  T^  in  the  members  AD,AF, 
FG  must  balance  the  external  forces  5  tons  and  4^  tons  in  order 
that  the  equilibrium  of  the  remainder  of  the  truss  may  be  pre- 
served. 

Hence,  revolving  horizontally  and  vertically, 

r,  +  r,  cos  («  +  30°)  -  C  sin  60°  =  o ; 
7;  sin  {a  +  30°)  -  C  cos  60°  +  5  -  4^  =  o. 
Taking  moments  about  F, 

Cs-  SBF  cos  {t,o'=  -  a)  -f  ^IDF  sin  30°  =  c. 


Rut 


3^3 


cos(«-f30°)  =  -4=-,  sin  («4-30°)=  ^-7^.  cos  (30°  -  a) =—^-. 
2 1^13  2  ^13  '2  Vii 


;!<; 

'M 


'  m 


ii 


lr:[ 


»^u  m 


f  I 


.  Ill 

r 

li 

}■ 

/; 

:i: 

\ 

9 

I 

jjl     !|1 

>  }■, . 

"     ' 

i  1'-' ' 

1    <; 

i  ^ 

■ 

i 

,1 

■ 

^4  THEORY  OF  STRUCTURES. 

BF—  BD  sec  a  =  5  ^Ti,      and   DF  =  5  tt. 


2  V13  2 


T-^.l^J-^-.^  +  i^zo. 


2  ^13 


<^".  5  -  5  .  5  v/13  . -4--  +  4i.  5  •i  =  O. 
2  V13 

Hence  C=  15^  tons,  Z,  =  9.89  tons,  and  7!,  ::=  6.35  tons. 

Ex.  2.  The  figure  represents  a  portion  of  a  bridge-truss  cut 

off  by  a  plane  A/N  and  supported  at 
the  abutment  at  A. 

Tiic  vertical  reaction  at  A 

=  409,400  lbs. 
^  "'f"^        The  weight  at  Z?  =  49,500  lbs. 
F,c.  ....     ^  «'         "        "   C:  =  38,700  lbs. 


409,400  11)S.  p, 


I*  ]| 


AB  =  BC=  24  ft. ;  BD  =  24  ft.;  {7/:  =  29^  ft. 

The  forces  C,  D\  T  in  the  members  met  by  MN  must 
balance  the  external  forces  at  A,  B,  C. 
Revolving  horizontally  and  vertically, 

T-\-  D'  cos  a  —  C  cos  /8  —  o; 

D'  sin  a  4-  C"'  sin  ft  —  409400  +  49500  +38700  =  o ; 

a  and    ft  being   the  inclinations  to  the  horizon  of  EF,  DEy 
respectively. 

Taking  moments  about  E, 

—  T  X  29^  4-  409400  X  48  —  49500  X  24  =  o. 


PIERS. 

2g\       II 


6^ 


But  tan  a  =  — ''-"  = 

24        9 


and     tan  /S  =  ^-  =  -. 
24       9 


.•.  sin  a  = 


II 


9 
cos  «  ^ ~ 


1/202 
Hence        T  =  62g,42y^j  lbs. ; 

981450O  ^/--       1090500 


,   sin/3=— ^,   cos /?  =  ~2_ 
4/202  1^85  ■v'ss 


c 


^r:f-*'«5  =  --^fp.^s 


lbs.; 


1994600    

/;  zz:  ■ r  202  lbs. 

27.  Piers. — 1  o  determine  the  stresses  in  the  members  of 
the  braced  piers  (Fig.  122)  supporting  a  deck  bridge. 


40  tons 


20  tons 


^2  tons//    9«! 

B 

f 

/ 

I 

'     E/            i 

y^ — 1— 

/ 

s 

G  H 

Fig.  132. 


Fic.  i»3. 


Z>rf/rt.— Height  of  pier  =  50  ft. ;  of  truss  —  30  ft.     Width 
>f  pier  at  top  :==  17  ft. ;  at  bottom  --  ii\  ft. 


T 

.1 

1  ' 

oo 


THEORY   OF  STRUCTURES. 


t    ! 


i 


The  bridge  w  hen  most  heavily  loaded  throws  a  weight  of 
l(X)  tons  on  eaeh  of  the  points  A  and  B. 

Weight  of  half-pier  =  30  tons. 

The  increased  weight  at  each  of  the  points  C,  D  and  E,  F, 
from  the  portions  AD  and  CF  oi  the  pier  =  5  tons. 

Resultant  horizontal  wind-pressure  on  train  =  40  tons  at 
87^  feet  above  base. 

Resultant  horizontal  wind-pressure  on  truss  =  20  tons  at 
65  feet  above  base. 

Resultant  horizontal  wind-pressure  on  pier  =  2|  tons  at 
each  of  the  points  C  and  E. 

With  the  wind-pressure  acting  as  in  the  figure,  the  diagonals 
CB,  ED,  and  GF  are  requireil.  When  the  wind  blows  on  the 
other  side,  the  \agonals  D  to  A,  F  to  C,  and  //  to  E  are 
brought  into  plaj-.  The  moment  of  the  couple  tending  to 
ove'  turn  the  pier 

=  40  X  87I  4-  20  X  O5  -1-  4  X  25  =  4900  ton-feet. 

33l 


The  moment  of  stability  =  {2(X)  -f-  30)  X 


3871!  ft.-tons. 


Thus  the  difference,  =  4900  —  3871^  =  1028^  ft.-tons,  must 

be  provided   for  in   the   anchorage.     The   pull   on   a   vertical 

10281^ 
anchorage-tie  at  6  =  — ,r-  =  SOA'r  tons. 
^  331'"' 

Again,  if  //  be  the  horizontal  force  upon  the  pier  at  A  due 

to  wind-pressure, 

//■  X  50  =  40  X  87A-  4-  20  X  65  =  4800  ; 
//  —  96  tons. 

The  stress  diagram  can  now  be  easily  drawn. 

The  reciprocals  of  the  points  A,  /),  C,  D,  E,  F  are  4321. 
2561,  11-10-4169,  65789,  13-12-11-98  14,  and  87-1 5-16- 14,  respec- 
tively. In  the  stress  diagram  43  =  96  tons,  32  =  25  =  100  tons, 
57  =  7-15  =  4-IO  =:  1 1-12  =  6  tons,  and  10-11=12-13  =  2^ 
tons.  The  stress  in  EG  is  of  an  opposite  kind  to  the  stresses 
in  AC,  CE. 


Note. 
braced  \ 
maxim  u 
pressure 
structure 
of  the  st 
posts  of 
the  bridt 


Dcscripi 


Hoarding  (| 
Hoarding  ai 
Cast-iron  pi 
Copper, 
Corru.  ,itf: 
I'clt.a.^i  I,  ■ 
Ffl;  and  gr.i 
Inilvanized  i 
l.allis  and  pi 
I'.intiles. . . . 
SiicL't  lead.. 
SlR-t--t-zinc.. 
Slieet-iron  (c 


Slicet-iron  (1 
Sliingler,  (i6- 

"  (lull 

.Sl)eathing  (i- 

"         (cli 

(asl 

Slates  (ordin; 


*^l'ites  (large) 


■'^iates  and  in 

Thatch 

Tiles 

Tiles  and  mo 

Timbering  ol 

roofs  (addit 


WEIGHT  OF  KOOF-COVERIXGS. 


67 


Note. — In  computin;4  the  stresses  in  the  leeward  posts  of  a 
l^raced  pier,  it  is  usual  in  American  practice  to  assume  that  the 
maximum  load  is  upon  the  bricl<^e  and  that  the  wind  exerts  a 
pressure  of  30  lbs.  per  sq.  ft.  upon  the  surfaces  of  the  train  and 
structure,  or  a  pressure  of  50  lbs.  per  sq.  ft.  upon  the  surface 
of  the  structure  alone.  The  negative  stresses  in  the  windward 
posts  of  the  pier  are  determined  when  the  minimum  load  is  on 
the  bridge,  the  wind-pressure  remaining  the  same. 


TABLE   OF    WEIGHTS   OF    ROOF-COVERINGS. 


Description  of  Coveririii. 


lioar'Jing  (J-inch) 

Hoarding  aiiiJ  sheet-iron.. 
Cast-iron  plates  (,§-inch).  . 

Copper  

Corru  .ito-l  Ik  ,1  and  laths. 

Felt,  a.-.;  li :*'i  ,d 

Kelt  and  gravel 

dalvanized  iron 

Laths  and  plaster 

i'.intiles 

Sheet  lead 

Sheet-zinc 

Sheet-iron  (corrugated). . . 


Sheet-iron  (16  W.G.)  and  laths 

Shingles  (i6inch) 

(h'lig) 


Sheathing  (i-inch  pine) 

"         (chestnut  and  maple) 
(ash,  hickory,  oak) 
Slates  (ordinary) 


tites  (large). 


Weight  of 
Covering 
in  lbs.  per 
sq.  ft.  of 
Covered 
Area. 


Slates  and  iron  laths 

Thatch 

Tiles 

Tiles  and  mortar 

Timbering  of   tiled   and    slate 
roofs  (additional) 


Dead  Weight  of  Roof  in  lbs.  per  sq. 
ft.  of  Covered  Area. 


2.5  to  3 

6.5 

15 

.8  to  1.25 

5-5 
.3  10  .4 

8  to  10 
I  to  3 

9  to   ID 

6  to  10 
5  to  8 

1.25  to  2 

3  4 

3-4 

5 
2 

3 

3 

4 

5 

5  to  () 


9  to  II 


10 

6.5 

7  to  20 

25  to  3c 
55  to  6.5 


S  without  boards  and  11  with  boards 

for  spans  u|)  to  75  ft. 
12  without  boards  and  15  with  boards 
for  spans  from  75  to  150  ft. 

ID  on  laths  for  spans  up  to  75  ft. 
14    on    laths    for    spans    from    75    to 
150  ft. 


13  without  boards  or  on  laths  and  16 

on   i^-in.  boards  for  s[)ans  up  to 

75  ft. 
17  without  boards  or  on  laths  and  20 

on  li-in.  boards  for  spans  from 

75  to  150  ft. 


;i;i 


.^: 


J! 

68 


THEORY  OF  STRUCTURES. 


WEIGHTS   OF    VARIOUS    ROOF-FRAMINGS. 


TABLE  O 


|i  I    - 


Description  of  Roof. 


Pent 

Common  Truss. 


^ 


Timber 
rafters  and 
struts,  iron 
ties 


Common  Truss. 


Location. 


j  Liverpool  I 
1     Dociis      ( 


j  Liverpool  » 
I     Docks      S 


Bowstring  ....   j Manchester 

"  Lime  Street 

"  Birmingham 

Arched Strasburg 

Paris 

Dublin 

Derby 

iSvdenham 


Cover- 
ing. 


Felt 


Zinc 

Zinc 
Zinc 
Slates 


St.  Pancras 
Cremorne 


Span 


Width 

of 
Bays.} 


Weight  in  ]bs. 

per  sq.  It.  of 

Covered  Area. 


Fram-     Cover- 
ing.   I     ing. 


ft.    in. [ft. 

15  o' 
37  o 
40  0 
50  o 


53  3 

54  o 

55  o 
72  o 


62  o 

76  o 

79  o 

80  8 

90  2 


84  0 
100  o 
,130  o 
'  50  o 

1154  o 
211 

97 
153 

41 

81 
120 

72  o 
240  o 

45 


5  <J 
12  o 
10  o 


3-5 
4.6 

5-5 
3-0 


II  o    2.085 

14  o    9.5 

6  6  1 1 . 6 

20  o    7.0 


12  o  3.013 

25  o  2.6 

13  o'  3.86 
II   S  a. 72 


5.00 


5.66 

7.72 

5-42 
12. 1 


20  o 


1.6 


9 
14 
26 
II 

21 

24 

13 

26 

16 

24 


00.5 

o    7.0 

o'  6.4 
II    9 . t ) 

"  4-9 
o  ii.o 
012.0 
o  15.0 


29 
14 


I 


10.7 
16.8 
fi.8 
II-3 
24.5 
II. 5 


Pitch. 


30" 


30" 

30° 

30° 

26°  34' 


TAE 


mm 


WIND   PRESSURES. 


69 


TABLE  OF  THE  VALtJES  OF"  Pn,  Pv ,  Ph,  IN  LBS.  PER  SQ.  FT.  OF 

SURFACE,   WHEN   />  =  40,  AS   DETERMINED   BY 

THE  FORMULA  Pn  =  P  .  sin  a>84  cosa-i. 


Pitch  of  Roof. 

Pn 

Pv 

Pk 

5° 

5.0 

4.9 

•4 

10' 

9-7 

9.6 

1-7 

20" 

18. 1 

17.0 

6.2 

3°: 

26.4 

22.8 

132 

40^ 

33-3 

25-5 

21.4 

50° 

38.1 

24-5 

29.2 

60' 

40,0 

20.0 

34-0 

70° 

41. c 

14.0 

38.5 

80' 

40.4 

7.0 

39-8 

90^ 

40.0 

0.0 

40.0 

TABLE    PREPARED   FROM   THE   FORMULA  /  = 


©■ 


Vc!ociiier>  in 

Velocities  in 

Pressurv;  in 

feet  per  second. 

miles  per  hour. 

lbs.  per  sq.  ft. 

10 

6.8 

•25 

20 

13.0 

I.OC 

40 

27.2 

4.CX3 

60 

40.8 

g.cx) 

70 

47.6 

12.25 

80 

54-4 

16.00 

90 

6r.2 

20.25 

100 

68.0 

25.00 

IIO 

74. 3 

30-25 

120 

81.6 

36.00 

130 

88.4 

42.25 

150 

102.0 

56.25 

\0 


'i   i  I'. 


70 


THEORY  OF  STRUCTURES. 


EXAMPLES. 

1.  Show  that  the  locus  of  the  poles  of  the  funicular  polygons  oi 
which  the  first  and  last  sides  pass  through  two  fixed  points  on  the  clus 
ing  line,  is  a  straigat  line  parallel  to  the  clcjsing  line. 

2.  The  first  and  last  sides  of  a  funicular  polygon  of  a  system  of  forces 
intersect  the  closing  line  in  two  fixed  points.  Show  that  for  any  position 
of  the  pole  each  side  of  the  polygon  will  pass  through  a  fixed  point  on 
the  closing  line. 

3.  Four  bars  of  equal  weight  and  length,  freely  articulated  at  the 
extremities,  form  a  square  ABCD.  Tiie  system  rests  in  a  vertical  plane, 
the  joint  A  being  fixed,  and  the  form  of  tlie  square  is  preserved  bv 
means  of  a  horizontal  string  connecting  the  joints  B  and  D.  If  \V  be 
the  weight  of  each  bar,  sliow  (ii)  tiiat  the  stress  at  C  is  horizontal  and 


W 


rVs 


=  — -,  (d)  that  the  stress  on  BC  at  B  is     IV12.  and  makes  with  tiie  ver- 
2  2 

tical  an  angle  tan  -'^,  (r)  that  the  stress  on  A/>  at  B  is      ll'J-lA  and 

makes  with   the   vertical   an   angle   tan"'ii,  (d)   that   the   stress   upon 
AB  at  A  is  |  PK  (c)  that  the  tension  of  the  string  is  2  IF. 

4.  Five  bars  of  equal   length  and  weight,  freely  articulated  at  tiie 

extremities,  form  a  regular  pentagon  ABCDli.     The  system  rests  in  a 

vertical  plane,  the  bar  CD  being  fixed  in  a  horizontal  position,  and  the 

form  of  the  pentagon  being  preserved  by  meani  of  a  string  connecting 

the  joints  B  and  E.     If  the  weight  of  each  bar  be    \V,  sliow  that  the 

\V 
tension  of  the  string  is  —   (tan  54""  +  3  tan   iS°),  and  find  the  niagni- 

2 

tudes  and  directions  of  the  stresses  at  the  joints. 

5.  Six  bars  (jf  equal  length  and  weight  (  =  W),  freely  articulated  at 
the  extremities,  form  a  regidar  hexagon  /IBCHEF. 

First,  if  the  system  hang  in  a  vertical  plane,  the  bar  AB  being  fixed 
in  a  horizontal  positicjn,  and  the  form  of  tlu-  hexagon  being  preser\'c  1 
by  means  of  a  string  connecting  the  middle  points  of  AB  and  DI-1,  siiow 

W 

and 


that  (li)  the  tension  of  the  string  is  3  \V,  (b)  the  stress  at  C  is 


-^3 


horizontal,  (r)  the  stress  at  D  is  W  if  ^^.  and  makes  with  the  vertical 

an  angle  cot  ■'  2  4/3. 

Second,  if  the  system  rest  in  a  vertical  plane,  t'^i'  bar  DF.  being  fixed 
in  a  horizontal  position,  aiul  tlie  form  of  tlie  hi;xagoii   being  ijreserved 


by  mean 
tii>:  tensi 

makes  w 

and  niak 

Thira 
and  the  f 
necling  . 

each  of  t 
AD  is  2  / 
at  the  joi 
from  the 

6.  She 
liorizonta 
which  are 

7.  If  t 
that  the  < 
spect  to  s 
which  is  i; 

8.  A  s> 
fixed  poinl 
the  joints, 
tangents  0 
are  in  ariti 

9.  Ifar 
librium  in 
angles  of  i 
iiig  from  tl 


10.  Thi 
The  systeii 
same  horiz 
the  stress  i 


II.  Thr 

tern  is  kept 

mine  the  st 

Wiiat  rt 


■'  m 


EXAMPLES. 


71 


hy  means  of  a  string  connecting  the  joints   C  and   F,   show   that   (<i) 

til.,  tension  of  the  string  is  IV  — --,  (b)  the  stress  at  C  is  IV  A/  V'-  and 

V3  *^    12 

makes  with  CB  a.n  angle  sin  -'i/_3_,  (f)  the  stress  at  B  is   PF/i/^. 

'124  "12 


and  mak  js  with  CD  an  angle  sin 


^   28 


Third,  if  the  system  hang  in  a  vertical  plane,  the  joint  A  being  fixed, 
and  the  form  of  the  hexagon  being  preserved  by  means  of  strings  con- 
necting A  wiih  the  Joints  E,  D,  and  C,  show  that  (a)  the  tension  of 
each  of  the  strings  AE  and  AC  is  IV  V'3 ,  (b)  the  tension  of  the  string 
All  is  2  \V\  and  determine  the  magnitudes  and  directions  of  the  stresses 
at  the  joints,  assuming  that  the  strings  are  connected  with  pins  distinct 
from  the  bars. 

6.  Show  that  the  stresses  at  C  and  F  in  the  first  case  of  Ex.  5  remain 
liDrizontal  when  the  bars  AF,  FE,  BC,  CD  are  replaced  by  any  others 
which  are  all  equally  inclined  to  the  horizon. 

7,  If  the  pole  of  a  funicular  polygon  describe  a  straight  line,  show 
tiiat  the  corresponding  sides  of  successive  funicular  polygons  with  re- 
spect to  successive  positii  s  of  the  pole  will  intersect  in  a  slr;iight  line 
which  is  parallel  to  the  locus  of  the.[)olc. 

8,  A  system  of  heavy  bars,  freely  articulated,  is  suspended  from  two 
^\\o.(\  points;  determine  the  magnitudes  and  <lirections  of  the  stresses  at 
the  joints.  If  the  bars  arc  all  of  equal  weight  and  length,  show  that  the 
tangents  of  the  angles  which  successive  bars  make  with  the  horizontal 
are  in  arithmetic  progression. 

9.  If  an  even  number  of  bars  of  equal  length  and  weight  rest  in  equi- 
librium in  the  form  of  an  arch,  and  ifdi,  <t.j,  .  .  .  a„  be  the  resjiective 
angles  of  inclination  to  the  horizon  of  the  ist,  2d,  .  .  .  «th  bars  count- 
ing from  the  top,  show  that 

2;/  -f  I 

tan  (V„  +  ,  = 


m 


2lt 


tan  a„. 


10.  Three  bars,  freely  articulated,  form  an  equilateral  triangle  ABC. 
The  system  rests  in  a  vertical  plane  upon  supports  at  B  and  C  in  the 
same  horizontal  line,  and  a  weight  H^  is  suspended  from  ^l.  Determine 
the  stress  in  BC,  neglecting  tlie  weight  of  the  bars. 

W 


■bis. 


Vi 


II.  Three  bars,  freely  articulated,  form  a  triangle  AliC,  and  the  sys- 
tem is  kept  in  equilibrium  by  three  forces  acting  on  the  joints.  Deter- 
mine the  stress  in  each  bar. 

What  relation  holds  between  the  stresses  when  the  lines  of  action  of 


PI 


72 


THEORY  OF  STRUCTURES. 


t 


V    ■' 


%  f. 


(S: 


the   forces   meet  (<?)   in   the   centroid,    (b)    in    the   orthocentre   of  the 
triangle  ? 

12.  A  triangular  truss  of  white  pine  consists  of  two  equal  rafters  AB, 
AC,  and  a  tie-beam  BC\  the  span  of  the  truss  is  30  ft.  and  its  rise  is  7i 
ft, ;  the  uniformly  distributed  load  upon  each  rafter  is  8400  lbs.  Deter- 
mine the  stresses  in  the  several  members. 

Alts.  Stress  in  EC  =  8400  lbs.,  in  AB  —  4200  I/5  lbs. 

13.  ABCD  is  a  quadrilateral  truss,  AB  and  CD  being  horizontal  and 
15  and  30  ft.  in  length,  respectively.  The  length  of  AC  is  10  ft.,  and  its 
inclination  to  the  vertical  is  60°.  A  weight  JFi  is  placed  at  C,  and  \Vi 
at  D.  What  must  be  the  relation  between  f/'i  and  IV-i  so  that  the  truss 
may  not  be  deformed?  For  any  other  relation  between  IV \  and  IVi, 
explain  how  you  would  modify  the  truss  to  prevent  deformation,  and 
find  the  stresses  in  all  the  members. 

Ans.    Wx  =  IV,.1ASL1. 

2 

14.  A  Warren  girder  80  ft.  long  is  formed  oifive  equilateral  triangles. 
Weights  of  2,  3,  4,  5,  tons  are  concentrated,  respectively,  at  the  ist,  2d, 
3d,  and  4th  apex  along  the  upper  chord.  Determine  the  stresses  in  all 
the  members  of  the  girder. 

Ans. —  Tension  Chord  :  Stress  in  ist  bay  =  2  ^3 ;  2d  =  5^  4/3 ; 

3d  =  7  1/3;  4th  =  6^4/3;  5th  =  2f  4/3- 
Compression  Chord  :  Stress  in  ist  bay  =  44/3  ; 

2d  =  624/3 ;  3d  =  7JV3 ;  4th  =  5.^i/"3. 
Diagonals :  Stress  in  ist  and  2d.  =44/3; 

3d  and  4th  =  2^  V3;  5th  and  6th  =  f  4/3  ; 

7th  and  8th  =  24/3;  9th  and  loth  =  5i  Vz- 

15.  In  a  quadrilateral  truss  ABCD,  AD  is  horizontal,  AB  and  BC  are 
inclined  at  angles  of  60°  and  30°  respectively  to  the  horizontal,  and  CD 
is  inclined  at  45"  to  the  horizontal.  What  weight  must  be  concentrated 
at  C  to  maintain  the  equilibrium  of  the  frame  under  a  weight  JFat  B? 

h  a  weight  IV  is  placed  at  Cas  well  as  at  D,  what  member  must  be 
introduced  to  prevent  distortion.'  What  will  be  the  stress  in  that 
member  ? 

Ans.  First:  W^^J^.. 

2 

Second:  Introduce  brace  BD  and  let  BDA  =  a. 

W{2  -  4/3) 


16.  T 
intermed 
into  six 
depth  of 
diagrams 
\a)  A 
ib)   W 


17.  Th 
respective 
points  of 
point  /)  0 
The  truss 
Find  the  s 
members  v 
lbs.  per  lin 

A> 


Then  stress  in  BD  — 


2  sin  (60°  -I-  a) 


18.  If  ii 
whole  of  tl 
stress  diagr 

19.  A  tr 
beam  BC,  a 
from  A  by  1 
each  rafter 
in)  the  stres 
What  {b)  wi 
liobeam  be 
tilted  again.' 
pressure  bet 

Ans.  {a)  Si 
(*) 


.  ■)' 


.SAMPLES. 


n 


i6.  The  boom  AB  of  tne  accompanying  truss  is  supported  at  fiv'e 
intermediate   points  dividing  the   lengt'i 


into  six  segments  each  lo  ft.  long, 
depth  of  the  truss  =  lo  ft.  ]_.  aw 
diagrams  for  the  following  cases  : 


The 
tress 


ib:;ofirf4f5/B 


Fig.  J24. 

(rt)  A  weight  of  100  lbs.  at  each  intermediate  point  of  support. 
{b)   Weights  of  100,  200,  300,  400,  500  lbs.       in  order  at  these  points. 
Ans.  (a)  Stress  in  «  =  375  ;  ^  =  325 ;  c—  375  ;  h  =  450 ; 
;«  =  125  1^13:  n-  soVS'  <}-  50^5; 
p  =  2i  Vii  lbs. 
(J))  Stress  in  «  =  875  ;  /^  =  825  ;  <r  =  925;  //  =1350;  ^=  1325, 

<?=  1 125;  7  =  1375;  w;  =  50  i^5;  «  =  looV's; 
o  =  141^  VTi ;  p  =  HVTt,  ■  r  =  200V5  ; 
J  =  250  Vs ;  /  =  458*  V'i'3  lbs. 

17.  The  rafters  y4^,  AC  oi  a  factory  roof  are  18  and  24  ft.  in  length 
respectiv^ely.  The  tie  BC  is  horizontal  and  30  ft.  long.  The  middle 
points  of  the  rafters  are  supported  by  struts  D£,  DF  from  the  middle 
point  D  of  the  tie  BC\  the  point  D  is  supported  by  the  tie-rod  AD. 
The  truss  carries  a  load  of  500  lbs.  at  each  of  the  points  E,  A,  and  F. 
Find  the  stresses  in  all  the  members.  Secondly,  find  the  stresses  in  the 
members  when  the  rafter  AB  is  subjected  to  a  normal  pressure  of  300 
lbs,  per  lineal  ft.,  rollers  being  at  C 

Ans.  Stress  in iJ^"  =  \\\'i\\  EA  =  Zqo\  CF=  1016^;  FA  =600; 
BD  =667^;  (7/^  =  813*  ;Z;£'=3i2i;Z>F=4i6f; 
AD  =  502  lbs. 
Stresses  due  to  300  lbs.  in  BE  =  1012^  ;  EA  =  1800; 
^1/^  =  2812^;  BD=3847i;  AD  =  22S0  ; 
DC  =  2160  ;  AC  =  2700  ;  DF  =  o. 

18.  If  it  be  assumed  in  the  first  part  of  the  last  question  that  the 
whole  of  the  weight  is  concentrated  at  the  points  E  and  F,  draw  the 
stress  diagram. 

19.  A  triangular  truss  consists  of  two  equal  rafters  AB,  AC  Sind  a  tie- 
beam  BC,  all  of  white  pine;  the  centre  D  of  the  tie-beam  is  supported 
from  A  by  a  wrought-iron  rod  AD;  the  uniformly  distributed  load  upon 
each  rafter  is  8400  lbs.,  and  upon  the  tie-beam  is  36000  lbs. ;  determine 
(ii)  the  stresses  in  the  different  members,  /?C  being  40  ft.  and  /ID  20  ft. 
What  {b)  will  be  the  effect  upon  the  several  members  if  the  centre  of  the 
lie-beam  be  supported  upon  a  wall,  and  if  for  the  rod  a  post  be  substi- 
tuted against  which  the  heads  of  the  rafters  can  rest.'  Assume  that  the 
pressure  between  the  rafter  .ind  post  acts  at  right  angles  to  the  rafter. 

Ans.  (a)  Stresses  in  BD  =  13200;  AD  —  18000;  AB  =  13200  4/2  lbs. 
{b)        "         "       =    4200;    "    =    8400;    "    =    6300  V'2~lbs. 


'U; 


1   41 


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74 


THEORY  OF  STRUCTURES. 


.    1 


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20.  A  triangular  truss  of  white  pine  consists  of  a  rafter  AC,  a  vertical 

post  A/>\  and  a  horizontal  tie-beam  BC;  the  load  upon  the  rafter  is  300 

lbs.  per  lineal  foot;  ^!C=  30  ft.,  AB  =  6  ft.     Find  the  resultant  pressure 

at  '  '. 

Ans.  4409  lbs. 

Find  the  stresses  in  the  several  members  when  the  centre  D  of  the 

rafter  is  also  supported  by  a  strut  from  B. 

Ans.  Stress  in  BC  =  4500  y6\  CD  =  22500;  DB  =  11250; 

J)A  =  1 1250;  .IB  —  2250  lbs, 

21.  The  rafters  y//>',  AC  oi  a  roof-truss  are  20  ft.  long,  and  are  sup- 
ported at  the  centres  by  the  struts  Uh',  DF\  the  centre  D  of  the  tie- 
beam  BC  is  supported  by  a  tie-rod  AD,  10  ft.  long;  the  uniformly  dis- 
tributed load  upon  AB  is  Sooo  lbs.,  and  upon  AC  is  2400  lbs.  Determine 
the  stresses  in  all  the  members. 

What  will  be  the  effect  upon  the  several  members  ii  AB  be  subjected 
to  a  horizontal  pressure  of  156  lbs.  per  lineal  foot? 

A/is.  {ii)  Stress  in  BD  —  4600  ^^3  ;  BE  =  9200;  EA  =  5200; 
£D  =  4000 ;  AD  —  2600 ;  DE  =  1200 ; 

AE  =  5200 ;  CE  =  6400  ;  CD  =  3200  |''3. 
(d)  Tens,  in  BE  =  520  v'3 ;  -ID  —  260  V'3  ;  compres.  in 

ED  =  520  i/3";  AC  =  520  4/3  ;  DC=  780. 
No  stresses  in  BD,  AE. 

22.  Determine  the  stresses  in  all  the  members  of  the  truss  in  the 
preceding  question,  assuming  the  tie-beam  to  be  also  loaded  with  a 
weight  of  600  lbs.  per  lineal  foot. 

Ans.  Stress  in  ^IB  iiicicased  by  6000  4/3  lbs. ;  in  BC  by  9000  lbs.; 
in  AD  by  6000  4/3  lbs. 

23.  A  horizontal  beam  is  trussed  and  supported  by  a  vertical  strut  at 
its  middle  point.  If  a  loaded  wheel  roll  across  tiie  beam,  show  that  the 
stress  in  each  member  increases  proportionately  with  the  distance  of  the 
wheel  from  the  end. 


IV 
^Lns.  Stress  in  tie-beam  (hor.)  =  -j--^'  ^ot  ^! 


on  tie  = 


JVx 
I  sin  0' 


on  strut  = 


W 


-2X. 


24.  A  frame  is  composed  of  a  horizontal  top-beam  40  ft.  long,  two 
vertical  struts  3  ft.  long,  and  three  tie-rods  of  which  the  middle  one  is 
horizontal  and  15  ft.  long.  Find  the  stresses  produced  in  the  several 
members  when  a  single  load  of  6000  lbs.  is  concentrated  at  the  head  ot 

each  strut. 

Ans.  Stress  in  horizontal  members  =  50000  lbs. 
"       "   sloping  "  =  51420    " 

"       "   struts  =  12000    " 


U:  " 


EXAMPLES. 


7S 


25.  If  a  wheel  loaded  with  i2ocx)  lbs.  travel  over  the  top-beam  in  the 
last  question,  what  membeis  must  be  introduced  to  prevent  distortion? 
What  are  the,  niaximum  stresses  to  which  these  members  will  be  sub- 
jected ? 

Ans.  191 22  lbs. 

26.  A  beam  of  30  ft.  span  is  supported  by  an  inverted  queen-truss, 
tiie  queens  beinij;  each  3  ft.  long  and  the  bottom  horizontal  member  10 
ft.  ioni;.  Find  the  stresses  in  the  several  members  due  to  a  weic,'ht  /P 
at  the  head  of  a  queen,  introducing  the  diagonal  required  to  prevent 


distortion.     Also  find  the  stresses  due  to  a  weight  JT'at  centre  ot  beam 
Ans. 


I.   Stress  in  AB  =  ~°ir-  AE  =  2.32  IF;  EF  =  =°]V; 
9  9 


BE=:     -II 

3 
CF  = 


HF:=\.\61V 
1)F=  i.\6lV. 


BC  . 


10 , 


IV\ 


2.  Stress  in  AB  =^  "W; 

3 

EF  =  ^  H-'. 


AE^  \.7\IV;  BE=. 


IV 


27.  A  roof-truss  of  20  ft.  span  and  8  ft.  rise  is  composed  of  two 
rafters  and  a  horizontal  tie-rod  between  the  feet.  The  load  upon  the 
truss  =  500  lbs.  per  foot  of  span.  Find  the  pull  on  the  tie.  What  would 
the  pull  be  if  the  rod  were  raised  4  ft.? 

A/is.  3125  lbs.;  6250  lbs. 

28.  The  rafters  AB,  AC  of  a  roof  are  unequal  in  length  and  are  in- 
clined at  angles  (x,  fi  to  the  vertical ;  the  uniformly  distributed  load  upon 
AJi  =  IVi ,  upon  ^IC  =  Jl'-2 .     Find  the  tension  on  the  tie-beam. 

JVi  +  11^     sin  (I    sin^i 
2  sin  {(I  +  fJY 


Alls. 


29.  In  the  last  question,  if  the  span  =  10  ft.,  a  =  60'  and  /i  =  45  \  find 
the  tension  on  the  tie,  the  rafters  being  spaced  2^  ft.  centre  to  centre, 
and  the  roof-load  being  20  lbs.  per  square  foot. 

Alts.  198  lbs. 

30.  The  equal  rafters  AB,  AC  for  a  roof  of  10  ft.  span  and  2^  ft.  rise 
are  spaced  2^  ft.  centre  to  centre ;  the  weight  of  the  rcjof-covering,  etc. 
=  20  lbs.  per  square  foot.  Find  the  vertical  pressure  auvl  outward  thrust 
at  the  foot  of  a  rafter. 

Alls.  Total  vertical  pressure  =  125^/5  lbs.  —  horizontal  thrust. 

31.  The  lengths  of  the  tie-beam  and  two  rafters  of  a  roof-truss  are  in 
the  ratios  of  5  :  4  :  3.  Find  the  stresses  in  the  several  members  when 
tlie  load  upon  each  raftt^r  is  uniformly  distributed  and  equal  to  100  lbs. 

.Ins.  Stress  in  tie  =  48  lbs.  ;  in  one  nifter  =  60  lbs.;  in  otlier  =  80  lbs. 


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THEORY  OF  STRUCTURES. 


32.  In  a  triangular  truss  the  rafters  each  slope  at  30° ;  the  load  upon 
the  apex  =  100  lbs.  Find  the  thrust  of  the  roof  and  the  stress  in  each 
rafter. 

Ans.  100  lbs.;  86.6  lbs. 

33.  A  roof-truss  is  composed  of  two  equal  rafters  and  a  tie-beam,  and 
the  span  =  4  times  the  rise;  the  load  at  the  apex  =  4000  lbs.  Find  the 
stresses  in  the  several  members. 

Secondly,  if  a  man  of  150  lbs.  stands  at  the  middle  of  a  rafter,  by 
how  much  will  the  stress  in  the  tie-beam  be  increased  ? 

Ans. — 1.  Stress  in  tie  =  4000  lbs. ;   in  each  rafter  =  2000  1/5  lbs. 
2.  75  lbs. 

34.  A  king-post  truss  for  a  roof  of  30  ft.  span  and  7i  ft.  rise  is  com- 
posed of  two  equal  rafters  AB,  AC,  the  horizontal  tie-beam  BC,  the 
vertical  tie  AD,  and  the  struts  DE,  DF  from  the  middle  point  D  of  the 
tie-beam  to  the  middle  points  of  the  rafters ;  the  roof-load  =  20  lbs.  per 
square  foot  of  roof-surface,  and  the  rafters  are  spaced  10  ft.  centre  to 
centre.     Find  the  stresses  in  the  several  members. 

Second,  find  the  altered  stresses  when  a  man  of  150  lbs.  weight  stands 
■on  the  ridge. 

Third,  find  the  altered  stresses  when  the  tie-beam  supports  a  ceiling 
weighing  12  lbs.  per  square  foot. 

Ans. — I.  Stress  in  BE  =  56250  lbs. ;  BD  —.  2250  i/Jlbs.; 
AE  =  '6875  lbs. ;  DE  =  9375  lbs.; 

AD  =  7  500  \/J  lbs. 

2.  Stresses  in  BD,  BE,  AE  increased  by  150  lbs.,  75  4/5 

lbs.,  and  75  ^5  lbs.,  respectively ;  other  stresses  un- 
changed. 

3.  Stresses  in  AD,  tie-beam,  and  rafters  increased  by  1800, 

1800,  and  900  -/s  lbs.,  respectively  ;  other  stresses 
unchanged. 

35.  The  platform  of  a  bridge  for  a  clear  span  of  60  ft.  is  carried  by 
two  queen-trusses  15  ft.  deep;  the  upper  horizontal  member  of  the  truss 
is  20  ft.  long;  the  load  upon  the  bridge  =  50 lbs.  per  square  foot  of  plat- 
form, which  is  12  ft.  wide.     Find  the  stresses  in  the  several  members. 

Ans.  Stress  in  vertical  --=  6000  lbs.;   in  each  sloping  member 
=  loooo  lbs. ;  in  each  horizontal  member  =  8000  lbs. 

36.  If  a  single  load  of  6000  lbs.  pass  over  the  bridge  in  the  last  ques- 
tion, and  if  its  effect  is  equally  divided  between  the  trusses,  find  (a)  the 
greatest  stress  in  the  members  of  the  truss,  and  also  {l>)  in  the  members 
which  must  be  introduced  to  prevent  distortion.  Also  find  (c)  the 
stresses  when  one  half  the  bridge  carries  an  additional  load  of  50  lbs.  per 
square  foot  of  platform. 


EXAMPLES. 


77 


Am. — {a)  In  sloping  end  strut  =  3333^^  lbs.;  horizontal  tie  = 
26668^  lbs.;  horizontal  strut  =  1333^  lbs. 

{c)  In  sloping  end  strut  =  6250  lbs.;  horizontal  tie  = 
5000  lbs.;  horizontal  strut  =  3000  lbs. 

{b)  In  case  (a)  =  i666f  lbs.;  in  case  {c)  =  2500  lbs. 

37.  A  roof-truss  consists  of  two  equal  rafters  AB,  AC  inclined  at  60° 
to  the  vertical,  of  a  horizontal  tie-beam  BC  of  length  /,  of  a  collar-beam 

DK  of  length  — ,  and  of    queen-posts   DF,  EG  at  each  end  of    the 

collar-beam  ;  the  truss  is  loaded  with  a  weighs  of  2600  lbs.  at  the  vertex, 
a  weight  of  4000  lbs.  at  one  collar-beam  joint,  a  weight  of  1200  lbs.  at 
the  other,  and  a  weight  of  1 500  lbs.  at  the  foot  of  each  queen  ;  the 
diagonal  DG  is  inserted  to  provide  for  the  unequal  distribution  of  load. 
Find  the  stresses  in  all  menioers. 

Alts.     Stress  in  BD  =  i i733i ;  BF  =  5866^  ^3 ;  DF=  1^,00; 
DA  =  2600 ;  DE  =  3633i  f^  ;  /;(;  =.-  1S66S  ; 

GC  =  4933i  Vi  ;  OE  =  2433^ ;  CE  =  9866^  ; 
AE  =  2600  lbs. 

38.  The  rafters  AB,  AC  are  supported  at  the  centres  by  the  struts 
DE,  BE;  the  centre  of  the  tie-beam  is  supported  by  the  tie  AD; 
BC  =  30  ft.,  AD=  7i  ft.  ;  the  load  upon  AB  is  4000  lbs.,  that  upon  AC 
1600  lbs.  r.nd  the  stresses  in  all  the  members.  By  an  accident  the 
strut /?£■  was  torn  away;  how  were  the  stresses  in  the  other  members 
affected  ? 

Ans. — Case  i ;  Stress  in  BE  =  2400  ^5  ;  ^D  =  4800 ; 

DE  =  1000  v'5  :  AE  =  1400  v^5  ; 
AE  =  1400  1/5  ;  DE  =  400  1^5  ; 
FC  =  1800  j/J";  DC  =  3600  lbs. 

Case  2 :  Stress  in  BA  —  1400  \/$  ;  BD  =  2800 ; 
AD  =  400 ;  AE  =  1400  4/5  ; 
EC  =:  1800  {/J;  DE  =  400  4/5"; 
DC  =  3600. 

39.  The  platform  of  a  bridge  for  a  clear  span  of  60  ft.  is  carried  by 
two  trusses  15  ft.  deep,  of  the  type  shown  by  the  c       n      f 
aiconipanying    diagram ;     the     load    upon   the    - 
bridge  is   50  lbs.  per  square  foot  of  platform, 
which  is   12  ft.  wide.     Fuul  the  stresses  in  the 
several  members.  ^'°-  "'• 

Afis.     Stress  in  BE  =  1 3500 ;  BG  =  6750  \/J;  EG  =  40CO  ; 

ED  =  13500;  GD=  2250  I  7;  GA  =4500  V'J; 
AD  =  9000  lbs. 


;t 


'ii'. 


^'\\ 


78 


THEORY  OF  STRUCTURES. 


\    '\ 


^   i\ 


40.  If  a  single  weight  of  2000  lbs.  pass  ove--  a  truss  similar  to  that 
shown  in  the  preceding  question,  find  the  stresses  in  the  several  members 
when  the  load  is  (i)  at  E,  (2)  at  D. 

Ans. — Lase  i:  Stress  in  BG  =  1500  |/J;  BE=  3000; 

EG  =  2000 ;  ED  =  3000 ; 

GD  =  1000  1^5  ;  AG  =  500  ^5  ; 

AH- soo\/J;  DN=o;  F//  =  o; 
DF  =  1000  ;  FC  =  1000 ; 

C//  =  500  1/5"  lbs. 
Case  2 :  Stress  in  BA  and  CA  =  1000  1/5  ; 

BD  and  DC  —  2000 ;  AD  =  2000  lbs., 
and  in  other  members  =  o. 

41.  A  white-pine  triangular  truss  consists  ot  two  rafters  AB,  AC,  of 
unequal  length,  and  a  tie-beam  BC.  A  vertical  wrought-iron  rod  from 
A,  10  ft.  long,  supports  the  tie-beam  at  a  point  D,  dividing  its  length 
into  the  segments  BD  =  10  ft.  and  CD  =  20  it.  The  load  upon  each 
rafter  is  300  lbs.  per  lineal  ft. ;  the  load  upon  the  tie-beam  is  18,000  lbs., 
uniformly  distributed.  Determine  the  stresses  in  the  several  members. 
Ans.     In  AB  =  9650  V2  lbs.  ;  AC=  4825  Vj  lbs.  ;  BD  =  CD  =  9650  lbs. 

42.  The  post  of  a  jib-crane  is  10  ft.  ;  the  weight  lifted  =  IV;  the  jib 
is  inclined  at  30°,  and  the  tie  at  60°,  to  the  vertical.  Find  (a)  the  stresses 
in  the  jib  and  tie,  and  also  the  B.  M.  at  the  foot  of  the  post. 

How  (/>)  will  these  stresses  be  modified  if  the  chain  has/our  falls,  and 
if  it  passes  to  the  chain-barrel  in  a  direction  bisecting  the  angle  between 
the  jib  and  tie? 

Ans.— {a)  Stress  in  tie  =  IV;  in  jib  =  fV  ^J.  g,  m  _  j  |/Jft,  tons. 
id)       "         "      =.87 IV;      •'      =i.S7lV. 

43.  An  ordinary  jib-crane  is  required  to  lift  a  weight  of  10  tons  at  a 
horizontal  distance  of  9  ft.  from  the  axis  of  the  post.  The  hanging  part 
of  tile  chain  is  in  four  falls;  the  jib  is  15  ft.  long,  and  the  top  of  the 
post  is  i6i  ft.  above  ground.  Find  the  stresses  in  the  jib  and  tie  when 
the  chain  passes  (i)  along  the  jib,  (2)  along  the  tie. 

The  post  turns  round  a  vertical  axis.  Find  the  direction  and  magni- 
mde  of  the  pressure  at  the  toe,  which  is  3  ft.  below  ground. 

Ans, — (i)  Stress  in  tie  =  32jL5  tons;  in  jib  =  iij|  tons. 


II 


(2) 


If        <i 


=  (^.BJlI.  —  2ij  tons  ;  in  jib  =  9,^  tons. 


t  ,m 


Pressure  on  toe  =  10  i/io  tons,  and  is  inclined  to  vertical  at  an  angle 


EXAMPLES. 


79 


44.  In  the  crane  represented  by  the  figure  AB  =  AC  =  a 
35  ft.;  />'€=  20  ft.;  BDz=2oit.\  the  weight  lilted  =  25  tons;  r'^^j:;;;^^^ 
AC  slopes  at  45°  ;    the  cliain  hangs  in  four  falls  and  passes  ^^  "' 
from  A  to  D.    Find  the  stresses  in  all  the  members  and  c        o 
the  upward  pull  at  D.                                                                      F'o-  »»«• 

Ans.     Stress  in  BC  =  26  ;  AC  =  47.6 ;  AB  =  28.4 ;  CD  =  32.8  tons. 
Vertical  pull  at  Z>  =  31.3  tons. 

45.  The  figure  represents  the  framing  of  an  hydraulic  crane.     AB=BD 

=  IJF  =  y-Y;  =  //A'  =  5  ft.;  A'(J  =  BC  =  2i  ft.    Find  the 

"«--X^r^      Stresses  in  the  members  of  the  crane  when  the  weight 

K  g'f   dV'a  (I  ton)  lifted  is  {a)  at  A  ;  {/>)  at  B ;  (c)  at  /).     Also  Uf)  find 

Fio.  la?.        the  stresses  when  there  is  an  additional  weight  of  ^  ton 

at  each  of  the  points  B,  D,  F,  and  G. 

Ans— {a)  Stress  in  tons  in  AB  -  BD  ='2  ;  DP  =  FG  =  -"  ; 

KG  =  L'/  ;  AC=    )/J;  CE  =  -  vT; 
20  9 

i^//=^=^=Vr;C/;=5_^- 
9009  9 

99  _  143 

NG  =  ^^;  BC=EF  =  o. 
26 


(S)  Stress  in  tons  inAB  =  oz=  AC;  BC—  i;  CE=^^  \'2 ; 


HE  =  ^  V7;  Z)F=  /-(;  =  — ;  GA'  =  -^; 
13  u  2 

CZ?=7V5';  DE=  ^V'JT?; 
9  99 


28 

II 

^  V^3"T7; 

7  i^r 


EC  = -^~  V i\7  ;  HG  ^        -. 
143  26 

(f)  Stress  in  ^^  =  o  =  AC  =  BC  =  DC  =  BD  ^  CE\ 

DF=^-^^  =  FG;  DE  =  ^^  \'^t, 

^^=.f3^^7''''^=2l*'5"; 


HE 


20    ,- 


=    -  |/2  ;  KG  =  2i. 
13 


!iC 


■i  I' 

4:   * 


mm 


m 


\''  •}  *  i 


^ 


fee 


80  THEORY  OF  STRUCTURES. 


id)  Stress  in  tons  in  AB  =  0  =  AC;  BC=-=  EF; 

21  62  5  \/z 

FD=  -  =  FG,  KC=-\  CE=-^ 
I'  13  9 


^^  =  3ov:2  ;/?,?  = 


J3 


8 
99 


^317; 


Fig.  ia8. 


46.  The  inclined  bars  of  the  trape- 
zoidal truss  represented  by  the  figure 
make  angles  of  45°  with  the  vertical ;  a 
load  of  10  tons  is  applied  at  the  top 
joint  of  the  left  rafter  in  a  direction  of 
45°  with  the  vertical.  Assuming  the 
reaction  at  tlie  right  to  be  vertical,  fintl 
the  stresses  in  all  the  pieces  of  the  frame. 


Ans.    Vert,  reaction  at  Z>  =   -  t/2  ;   stress  in  DE  =  ■—  \'2\ 

•J  J 

'  DB  =  6i;  BE  =  6^;  BA  =  "~  ^2  ;  AE  =  -  s/z; 


AC 


CE  =  -  V2  tons. 


10 

;  CA  = 
3  3 

47.  The  post  of  a  derrick-crane  is  30  ft.  high  ;  the  horizontal  traces  of 
the  two  back-stays  are  at  riglit  angles  to  each  other,  and  are  15  ft.  and 
25  ft.  in  length.  Show  that  tiie  angle  between  the  shorter  trace  and  the 
plane  of  the  jib  and  tie,  when  the  stress  in  the  post  is  a  maximum,  is 

30'  58'. 

Also  find  the  greatest  stresses  in  the  different  members  of  the  crane 
when  the  jib,  which  is  50  ft.  long  and  is  hinged  at  the  foot  of  the  post,  is 
inclined  at  45°  to  the  vertical,  the  weight  lifted  being  4000  lbs. 

Ans.  Stress  in  jib  =  66661}  lbs.;  in  tie  =  4768.4  lbs.;  max. 
thrust  along  post  =  10991.5  lbs.  ;  max.  stress  on  Ions.; 
back-stay  =  7362.7  lbs.;  on  short  back-stay  =  10539  'bs. 

48.  A  queen-truss  for  a  roof  consists  of  two  horizontal  members,  the 
lower  48  ft.  long,  the  upper  16  ft.  long;  two  inclined  members  AB,  DC, 
and  two  queens  BE,  CF,  each  8  ft.  long;  the  points  if, /"divide .,4Z>  into 
three  equal  segments ;  the  load  upon  the  members  AB,  BC,  CD  is 
120  lbs.  per  lineal  foot.  Find  (a)  tlie  stresses  in  the  several  members. 
How  {b)  will  these  stresses  be  modified  if  struts  are  introduced  from  the 


m 


EXAMPLES. 


8l 


feet  of  the  queens  to  the  middle  points  G,  H  of  the  inclined  members  ? 
In  this  latter  case  also,  determine  (r)  the  stresses  due  to  a  wind-pressure 
of  1 20  lbs.  per  lineal  ft.  normal  to  AB,  assuming  that  the  horizontal  re- 
action is  equally  divided  between  the  two  supports  at  A  and  D. 

Ans.—{a)  Stress  in  lbs.  in  AE  =  4066.56  =  EF  =  DE  =  BC\ 
AB  =  4546.  yb^CD;  BE  =  2033. 28  =  CF. 
(b)  Stress  in  lbs.  in  AE=  5' 39- ^4  =  E>E\ 

BC  =  4066. 56  =  AE\  AG  =  5746. 56  rr-  n//; 
BG  =  4546.56  =  C7/;  EG  -  1 200  =  EH; 
BE  =  536.64  ~  CE. 
(c)  Additional  stress  in  AG  =  1040  f'J ;  BG  =  680  4,^5  ; 
GE=  600  v^5 ;  AE=2^2o  ;  BE  =  600  ;  BC  —  400  \^$  i 
BE=  400  4/5  ;  CE  =  400  ;  CB  =  400  {/$  ;  EE  =  1 1 20 ; 
/^Z>=320. 
(In  case  (c)  the  brace  BE  is  introduced  to  prevent  distortion.) 

49.  A  pair  of  shear-leg.s,  each  25  ft.  long,  with  the  point  of  suspension 
20  ft.  vertically  above  the  ground  surface,  is  sinnnorted  by  a  tie  100  ft. 
long;  distance  between  feet  of  legs  =  10  4/5  ft.  Find  the  thrusts  along 
the  legs  and  the  tension  in  the  tie  when  a  weight  of  2  tons  is  being 
lifted. 

Ans.  Tension  in  tie  =  1. 137 tons;  compn.  in  each  leg  =  1.87  tons. 

50.  In  the  crane  ABC,  the  vertical  post  AB  =  15',  the  jib  AC=  23', 
iir;!  the  angle  BAC  =  30".  Find  (a)  the  stresses  in  the  jib  and  tie,  and 
also  the  bending  moment  at  the  foot  of  the  post  when  the  crane  lifts  a 
weight  of  4  tons. 

The  throw  is  increased  by  adding  two  horizontal  members  CE,  BD 
and  an  inclined  member  DE,  the  figure  BE  being  a  parallelogram  and 
the  diagonal  CZ>  coincident  in  direction  with  CA.  Find  {b)  the  stresses 
in  the  several  members  of  the  crane  as  thus  modified,  the  weight  lifted 
being  the  same. 

In  the  latter  case  show  {c)  how  the  stresses  in  the  members  are  affected 
when  the  chain,  which  is  in  four  falls,  passes  from  E  to  //  and  then  down 
the  post. 

Ans. — (rt)  Tension  in  tie  =  3!  tons;    thrust  in  jib  =  6/j  tons; 

(b)  Stress  in  CE  =  934;  in  ED  =  10.16;  in  CB  =■  13.49; 

in  CD  =  6.15;  in  DA  —  10.7  ;    in  BD  —  7  tons. 

{c)  Stress  in  CE  —  8,9;    in  ED  =  10.7;    in    CB  =  12.9; 

in  CD  =  5.8  ;  in  DA  =  10.7  ;  in  BD  =  7.4  tons. 

51.  The  horizontal  traces  of  the  two  back-stays  of  a  derrick-crane 

are  .1  and  j  feet  in  length,  and  the  angle  between  them  is  ft     Show  that 

cos  (fi  —  0)      X 
the  stress  in  the  post  is  a  maximum  when  — :::,.,;  ^  •"=-.»  being  the 


cos  0 


angle  between  the  trace  x  and  the  plane  of  the  jib  and  tie. 


1 


.  iii»' 


m 


iir 


iiiJ 


if  'i 


83 


THEORY  OF  STRUCTURES. 


1 

., 

li 

1  )M 

' 

i 

3 

t 

- 

m 

■4 

.  ;■     1 

^  ''  ! 
.  i    ■ 

52.  The  two  back-stays  of  a  derrick-crane  are  each  38'  long,  and  the 
angle  between  their  horizontal  traces  2  tan"\'j;  height  of  the  crane- 
post  =  32';  the  length  of  the  jib  =  40';  the  throw  of  the  crane  =  20'; 
the  weight  lifted  =  4  tons.  Determine  the  stresses  in  the  several  mem- 
bers and  the  upward  pull  at  the  foot  of  each  back-stay  when  the  plane  of 
tlie  jib  and  post  (a)  bisects  the  angle  between  the  horizontal  traces  of 
the  back-stays,  (d)  passes  through  a  back-stay. 

Ans.  In  jib  =  5  ;  in  tie  =  2.52  tons  ;  in  back-stay  in  (a)  =  2.56, 
in  (d)  =  4.7  tons. 

53.  Find   the   stresses   in   the   members   of  the   crane   represented 
E  35'        B  ^y    ^''^    figure ;    also    find    balance-weight 

at  C. 

Ans.    Stress  in  BE  =  25  ;  DE  =  26.9; 
DB  =  21.08;     ZJA  =  26.08; 
BA  =  .24;  BC  =  18.12  tons. 
A  ~  Counterweight  at  C  =  15.14 

Fig.  I2Q.  tons. 

54.  Draw  tlie  stress  diagram  for  the  truss  represented  by  the  figure, 
the  load  at  each  of  the  points  B  and  C  being  500  lbs. 


angle  EA£>  =  3o\) 


00  V 


Fig.  130. 

Also,  if  the  rafter  AB  is  subjected  to  a  nominal  wind-pressure  of  100 
lbs.  per  lineal  ft.,  introduce  the  additional  member  required  to  prevent 
deformation,  and  state  in  lbs.  the  stress  it  should  be  designed  to  bear. 
Draw  the  stress  diagram  of  the  modified  truss,  assuming  that  the  foot  A 
is  fixed,  and  that  there  are  rollers  at  />. 
(AB  =  AE  =z  is' :  BC  =  10' ;  angle  BAD  =  45 

55.  The  post  AB  of  a  jib-crane  is  b 
20  ft.  ;  the  jib  AC  is  inclined  at  30°  and 
'lie  tie  BC  at  45°  to  the  vertical ;  the 
weight  lifted  is  5  tons.  Find  the 
stresses  in  the  jib  and  tie  when  the 
chain  passes  (a)  along  the  jib,  {d)  along 
the  tie,  (c)  horizontally  from  C  to  the 
post. 

The  chain  lias  /zoo  falls.  Fig.  i-ji. 

56.  In  a  mansard  roof  of  12  ft.  rise,  the  upper  triangular  portion  (of 
4  ft.  rise)  has  its  rafters  inclined  at  60°  to  the  vertical.     The  rafters  of  the 


"A" 

/B0\ 


A--Xii_/^jy 


c- 

&  tons 


^^55^^ 


EXAMPLES. 


«3 


lower  portion  are  inclined  at  30°  to  the  vertical.  If  there  is  a  load  of 
1000  lbs.  at  the  ridge,  find  the  load  at  each  intermediate  joint  necessary 
for  equilibrium,  and  the  thrust  of  tiie  roof. 

A  load  of  2000  lbs.  is  concentrated  at  each  of  the  intermediate  joints 
and  a  brace  is  inserted  between  these  joints.  Find  the  stress  in  the 
brace. 

Alls.   1000  lbs. ;  thrust  =  500  4/3  lbs.;  333^  4/ J  lbs. 

57.  The  horizontal  boom  CD  is  divided  into  eight  segments,  each 
8  ft.  long,  by  seven  intermediate  supports ; 

the  depth  of  the  truss  at  each  end  =  16  ft.; 
a  weight  of  i  ton  is  concentrated  at  C  and 
at  D.  and  a  weight  of  2  tons  at  each  of  the 
points  of  division.  Determine  the  stresses 
ill  tiie  several  members. 

58.  The  figure  is  a  skeleton  diagram  of  a  roof-truss  of  72  ft.  span  and 
12  ft.  deep ;  G,  K,  L,  O,  H  are  respectively  the  middle  points  of  AE,  EL, 

EF,  LF,  FB\  AE=EL^  LF=  FB  =  20  ft.; 
thetrussesare  12  ft.  centre  to  centre  ;  thedead 
weight  of  the  roof  =  12  lbs.  per  sq.  ft.;  the 
normal  wind-pressure  upon  AE  may  be  taken 
=  30  lbs.  per  sq.  ft.;  the  end  A  is  fixed  and 
Show  by  dotted  lines  how  the 


Fig.  13a. 


^>l     F   G    H  [I*' 


Fig.  133. 

B  is  on  rollers.    Draw  a  stress  diagram 

stress  diagram  is  modified  with  rollers  under  A,  B  being  fixed. 

59.  The  platform  of  a  bridge  of  84  ft.  span  bode 
nnd  9  ft.  deep  is  carried  by  a  pair  of  trusses  of 
the  type  shown  in  the  figure.     If  the  load  borne 
by  each  truss  is  300  lbs.   per  lineal  ft.,  find  the  Fig.  134. 
stresses  in  all  the  members. 

Ans.     Stress  in  AB  —  6000  ;  AC  =  1200  4/73  ;  AD  =  3600  |/i7 ; 
BC  =  4800 ;  CD  =  14400  ;  DE  =  28800. 
Stress  in  horiz.  chord  =  288000 ;  in  each  vertical  =  3600 
lbs. 

60.  The  figure  represents  the  shore  portion  of  one  of  the  trusses  for 

a  cantilever  higliway  bridge.  The  depth  of 
truss  over  pier  =  51  ft.;  the  length  of  each 
pane!  =  17  ft.;  the  load  at  A  (from  weight  of 
centre  span)  =z  16800  lbs.;  the  width  of  road- 
way =  15  ft.;  the  load  per  sq.  ft.  of  roadway 

P'°*  ^35-  =  80  lbs.     Find  the  stresses  in  all  the  mem- 

bers, assuming  the  reaction  at  the  pier  F  to  be  vertical. 

Ans.  /i  =  /j  =  28000 ;  /a  =  36500 ;  ^4  =  45000 ;  h  =  53500 ; 

/a  =  55200;  A  =48400;  /e  =  41600  =  /» ;    ci  =  5600  4/34; 

f,  =  7300  V34  ;  7^1  =  10200 ;  V3  =  15300;  Vt  =  20400  ; 


Afi    t«   h   ti  ti    tn  tr  ti  toB 


I'J.WW  lbs, 


'I 


84 


THEORY  OF  STRUCTURES. 


V\  =  25500;  Vt  =  45900;  V,  =  20400;  7/,  =  15300; 

v»  =  10200 ;   ca  =  9000  v^34  ;   r,  =  10700  t/34  ; 

Ci  =  \2\00  \^2>-^\   re  — 77500:    ri  =69000;  r»  =  60500; 

Ca  =  52000 ;  t/i  =  1700  V'34 ;  (ft  =  1700  V6i  ; 

r/i  =  1700  4/106  ;  «'<  =  22100;  (it.  =  1700  4/97  ; 

rt'o  =  3400  4/13  ;  ^A  =  8500  lbs. 

61.  The  inner  flange  of  a  bent  crane  forms  a  quadrant  of  a  circle  of 
20  ft.  radius,  and  is  divided  into /(3;/r  equal  bays.  Tlie  outer  flange 
forms  the  segment  of  a  circle  of  23  ft.  radius.  The  two  flanges  are  5  ft. 
apart  at  the  foot,  and  are  struck  from  centres  in  the  same  horizontal 
line.  The  bracing  consists  of  a  series  of  isosceles  triangles,  of  which  the 
bases  are  the  equal  bays  of  the  inner  flange.  The  crane  is  required  to 
lift  a  weight  of  10  tons.     Determine  the  stresses  in  all  the  members. 

62.  A  braced  semi-arch  is  10  ft.  deep  at  the  wall  and  projects  40  ft. 
The  upper  flange  is  horizontal,  is  divided  into/(3Kr  equal  bays,  and  carries 
a  uniformly  distributed  load  of  40  tons.  The  lower  flange  forms  the 
segment  of  a  circle  of  104  ft.  radius.  The  bracing  consists  of  a  series  of 
isosceles  triangles  of  which  the  bases  are  the  equal  bays  of  the  upper 
flange.     Determine  the  stresses  in  all  the  members. 

63.  The  domed  roof  of  a  gas-holder  for  a  clear  span  of  80  ft.  is  strength- 
ened by  secondary  and  primary  trussing  as  in  the  figure.    The  points  />' 

and  C  are  connected  by  the  tie  BPC  passing 
beneath  the  central  strut  AP,vih'\ch  is  15  ft. 
long,  and  is  also  common  to  all  the  primary 
trvisses ;  the  rise  of  A  above  the  horizontal  is  5 
ft.;  the  secondary  truss  ABEF  consists  of  the 
equal  bays  A//,  HG,  GB,  the  ties  BE,  EF,  FA,  of  which  BE  is  horizon- 
tal, and  the  stu-  .'.s  GE,  FH,  which  are  each  2  ft.  6  in.  long  and  are  par- 
allel to  the  radius  to  the  centre  of  GH\  the  secondary  truss  ACLK  is 
similar  to  ABEF;  when  the  holder  is  empty  the  weight  supported  by 
the  truss  is  36000  lbs.,  which  may  be  assumed  to  be  concentrated  at  G. 
H,  A,  M,  N,  in  the  proportions  8000,  4000,  1000,  4000,  and  8000  lbs.,  re- 
spectively.   Determine  the  stresses  in  the  different  members  of  the  truss. 

64.  The  flgure  is  the  skeleton  diagram  of  a  cantilever  for  a  viaduct  in 


4S  tgu  48  tdu  411  Ufa  JD  tou 


Fig.  137. 


EXAMPLES. 


85 


India.     Determine ^<,'-m/)///(((i//>'  the  stresses  in  the  various  members  un- 
der ilie  loading  indicated. 

f)5.  In  the  accompanying  roof-truss  AB  =  AC-=  30  ft.,  and  the  struts 
are  ail  normal  to  tlie  rafters.  Find  the 
stresses  in  all  the  members,  the  load  at 
each  of  the  joints  in  tlic  rafters  being  2 
tons  (angle  AliC  =  30'  and  angle  DEC 
=  10  ).  How  will  the  stresses  be  mod- 
ified if  theie  is  a  force  of  2  tons  acting 
at  each  of  the  points  of  support  between 
//  and  /)'  at  right  angles  to  the  rafter,  and  a  force  of  i  ton  at  A,  assum- 
ing tiiat  the  end  //  is  fixed  and  that  C  rests  upon  rollers? 

66.  The  figure  represents  a  portion  of  a  Warren  girder  cut  off  by  the 
1^  plane  iJAV  and  supported  upon  the  abutment  at 

j  A.    The  reaction  at  A  =  20  tons ;  the  load  con- 

centrated at  each  of  the  points  B  =  4.  tons.    Find 
the  stresses  in  each  of  the  members  met  by  AfN. 

Ans.  Stress  in  tension  chord  =  —  1/3  tons ; 


Fir..  138. 


Fig.  139. 


16 


.7  toni 


in  compression  chord  =  32  4/3  tons;  compression  in  diagonal  =  —  4/3 

tons. 

67.  The  figure  represents  a  portion  of  a  roof-truss  cut  off  by  a  plane 
MX  and  supported  at  A.  The  strut  DC  is 
vertical ;  AD  =  23  ft.,  and  the  distance  of  D 
from  AC=  7^  ft.;  the  angle  between  AC  and 
the  horizontal  r=  cos"'^  ;  the  vertical  reac- 
tion at  ^  =  7  tons ;  the  horizontal  reaction 
at  A  =  2i  tons ;  at  each  of  the  points  B  and 
Ca  weight  of  4  tons  is  concentrated.  Find 
the  stresses  in  the  members  met  by  MN. 
{AD  and  Ti  make  equal  angles  with  the 
rafter.) 

Ans.  Ci  =  13.2  tons ;   Tj  =  2.1  tons ;  Ti  =  10.8  tons. 

68.  The  feet  of  the  equal  roof-rafters  AB,  AC  are  tied  by  rods  BD, 
CD  which  meet  under  the  vertex  and  are  joined  to  it  by  a  rod  AD.  If 
H'l ,  IVi  are  the  uniformly  distributed  loads  in  pounds  upon  AB,  AC, 
respectively,  and  if  .S'  is  the  span  of  the  roof  in  feet,  find  the  weight  of 
metal  (wrought- iron)  in  the  ties. 

5     IVx  +  IV^ 


Pig.  140. 


Ans. 


—  S  cot  /3,    /  being  inch-stress  in  pounds,  and 


6  / 

fi  the  angle  ABD. 

{a)  If  AB  =  AC  =  20  ft..  AD  —  5  ft.,  the  angle  BAD  =  60°,  find 


'■i^M 


\   '  m 


Li 


'  n 


f  u 


fli 


mf  THEORY  OF  STRUCTURES. 

the  stresses  in  the  several  members  when  a  weight  of  3500  lbs.  is  con- 
centrated at  the  vertex. 

Ans.  7000  lbs.;  6309.8  lbs.;  3500  lbs. 

(d)  The  roof  in  (a)  is  loaded  with  10  lbs.  per  square  foot  on  one  side 
and  33  lbs.  per  square  foot  on  the  other;  the  trusses  being  13  ft.  centre 
to  centre.  Determine  (a)  the  stresses  ;n  the  several  members.  Examine 
(/>)  the  effect  of  a  horizontal  pressure  of  14  lbs.  per  square  fool  on  the 
most  heavily  loaded  side,  assuming  that  the  reaction  is  equally  divided 
between  th^  two  supports. 

A/is.   (a)  1 1 180  lbs.;   10077.65  lbs.;  5590  lbs. 
In   the  truss  represented  in  the  accompanying  figure,  the  load 
on  An  =  IV,  .  on  AC  =  IV^ ;  the  angle  ABD  =  H; 
AD  z=  BD=  AE  =  CE.      Find   the   total   weight 
of  metal  (wroiiglit-iron)  in  the  tie-rods. 


69. 


Fig.  141. 


Ans.    - 
6 


5     /F.  +  W; 


f 


Sco\.  /?;     5  being  the  span 


and /the  inch  stress. 

(a)  If  the  stress  in  BD  or  EC  is  equal  to  the  stress  in  DE,  show  that 

a 

/3  =  60° ;  a  being  the  angle  ABC. 

3 

{b)  The  trusses  are  12  ft. centre  to  centre;  the  span  is  40  ft.;  the  hori- 
zontal tie  is  16  ft.  long;  the  rafters  are  inclined  at  60°  to  the  vertical;  the 
dead  weight  of  the  roof,  including  snow,  is  estimated  at  10  lbs.  per  sq.  ft. 
of  roof-surface.  Determine  the  stress  in  each  member  when  a  wind 
blows  on  one  side  with  a  force  of  30  lbs.  per  sq.  ft.  normal  to  the  roof- 
surface,  assuming  that  the  horizontal  reaction  is  equally  divided  between 
the  supports. 

Ans.  Stress  in  AB  =  8956.8  lbs. ;  BD  =  10015.2  lbs.  =  EC; 
AD  =  2503.8  lbs.  =  AE;  DE  =  8196  lbs.; 
AC  =  11356.8  lbs. 

70.  In  the  truss  represented  by  the  accompanying  figure,  the  load 
upon  AB  =  JV,  ,  upon  AC  =  fF, ;  the  angle  ABD=z 
/3 ;  the  span  BC  =  S ;  the  ties  AD,  BD,  AE,  CE  are 
equal ;  E  and  G  are  the  middle  points  of  the  rafters. 
Find  the  amount  of  metal  in  the  tie-rods  (wrought- 
iron). 

5    ^ 
/ 


Ans.  -    — 


Fig.  142. 
IVi  +(JVi  +  JV,)  cos'  /3 


sin  /3    cos  /i 

(a)  The  struts  DE  and  EG  are  each  5  ft.;  the  angle  ABC  =  30° ;  the 
dead  weight  of  the  roof,  including  snow.  Is  9  lbs.  per  square  foot  of  roof- 
surface,  and  the  trusses  are  12  ft.  centre  to  centre.  Determine  the 
stresses  in  the  several  members  wlien  a  wind  blows  with  a  force  of  30 


EXAMPLES. 


87 


lbs.  per  square  foot  of  roof-surface  normal  to  the  side  AB.    The  span 
—  60  ft.,  and  the  (mkI  C  rests  upon  rollers. 

Secondly,  dctermirie  the  stresses  produced  in  the  members  of  the 
truss  in  the  preceding  question  when  a  single  weight  of  3000  lbs.  is  sus- 
pended from  G. 

//;«.— (I)  Stresses  in       DD\  DA;         DE\        EA ;        EC; 

31238.55;  19852.35;  12633.6;  8113.5;  24379.43; 
JiF;  FA;         FD;  CG;  GA;  GE. 

29620.44;  28685. 16  ;  7855.2  ;    22420.44;    21485.16;  1620  lbs. 


(2)  Stresses  in   BD; 

375  s'y) ; 

FA; 

2625; 


BE; 

2625; 


DA;         DE;         EA;  EC; 

4/^-  loooVT;  8751/39;  "25*^39: 
CG;         GA;  GE.  _ 

7875;       6375;        15001^3  lbs. 


'25  V39 
FD; 

o; 

(b)  The  rafters  AB,  AC  are  of  unequal  length  and  make  angles  of 
60°  and  45°,  respectively,  with  the  vertical;  the  strut  DF  =  7^  ft.;  the 
tie  DE  is  horizontal j  the  dead  load  upon  each  rafter  =  100  lbs.  per 
lineal  foot;  the  wind-pressure  normal  to  iB  =  300  lbs.  per  lineal  foot; 
rollers  are  placed  at  C.  Find  the  stresses  in  all  the  members.  The 
rafter  AB  =  45  ft. 

Show  by  dotted  lines  how  tlie  stress  diagram  will  be  modified: 

(1)  If  the  rollers  aie  placed  at  B. 

(2)  If  the  strut  DF  is  omitted. 

(3)  If  a  single  weight  of  500  lbs.  is  concentrated  at  D. 

(c)  If  it  is  assumed  that  the  horizontal  .eaction  is  equally  divided  be- 
tween B  and  C,  show  that  the  stress  in  DE  due  to  a  horizontal  wind- 
pressure  upon  AB  is  nil  ;  the  angle  ABC hc'xn^  30°. 

((/)  In  a  given  roof,  the  rafters  are  of  pitch-pine,  the  tie-rods  of 
wrought-iron  ;  the  span  is  60  ft.;  the  trusses  are  12  ft.  centre  to  centre; 
DF=  5  lx..=EG;  the  angle  ABC  =  30";  the  dead  weight  of  the  roof,  in- 
cluding snow,  is  9  lbs.  per  sq.  ft.  of  r'^'~ '-surface  ;  rollers  are  placed  at  C; 
a  single  weight  of  3000  lbs.  is  suspended  from  F,  and  the  roof  is  also 
designed  to  resist  a  normal  wind-pressure  of  26.4  lbs.  per  sq.  ft.  of  roof- 
surface  on  one  side  AB.  Determine  the  stresses  in  the  several 
members. 

71.  In  the  truss  represented  in  the  accompanying  figure,  the  struts 
DF.  DH,  EG,  EK  are  equal,  and  the  ties  BD,  AD,  a 

E.l,  EC  are  also  equal;  the  load  upon  AB  is  ff  1 ,     .    h^^^^k 
and  upon  AC  is  IV-, .      Find  the  weight  of  metal    -^^i^*^       ^"^ 


(wrought-iron)  in  the  ties. 


Ans.    -2- 


5     5    4^V^+j(lV,  + 


Fig.  143. 

fF,)cos'/S 


18  /  cos  fi   sin  fi 

(a)  AD  =  AE  —  BD  =1  EC  =  2-^  ft.;  the  angle  ABC  =  30° ;  the  span 
=  79  ft.;  the  trusses  are  13  ft.  centre  to  centre;  the  heel  B  is  free  to 


i.t  i 


88 


THEORY  OF  STRUCTURES. 


slide  on  a  smooth  wall-plate;  the  dead  weight  of  the  roof,  including 
snow,  is  8  lbs.  per  square  foot  of  roof-surface.  Determine  the  stress  to 
which  each  member  is  subjected  when  the  wind  blows  horizontally  with 
a  force  of  40  lbs.  per  square  foot  of  vertical  surface  (i)  upon  the  side 
AB,  (2)  upon  the  side  AC. 

Ans.  See  Ex.  3,  Art.  24. 
ib)  The  rafters  AB,  AC  are  inclined  at  60°  to  the  vertical  and  are 

each  40  ft.  in  length.  The  foot  Crests 
on  rollers,  and  the  foot  B  is  fixed.  The 
strut  DF  is  vertical,  is  10  ft.  long,  and 
is  equal  to  the  strut  DE  in  length. 
Also  AF  =  HF  =  10  ft.  The  dead  load 
carried  by  the  rafters  is  120  lbs.  per 
lineal  foot.  Provision  has  also  to  be 
made  for  a  normal  wind-pressure  upon 
AB  of  300  lbs.  per  lineal  foot.  Draw 
the  stress  diagram,  and  show  how  it 
will  be  modified  if  the  strut  DF  is  re- 
P'°-  '«•  moved. 

Ans.  Vertical  reaction  at  B  =  10528  lbs.  both  before  and  after 
DF  is  removed. 
Horizontal  reaction  at  ^  =  6000  lbs. 
The  dotted  lines  show  the  modified  stresses  for  one  half 
of  the  truss. 

72.  The  load  upon  a  roof-truss  of  the  accompanying  type  is  1000  lbs.  at 
each  joint ;  the  span  100  ft.;  the  rise  =  25  ft.    Find  the  stresses  in 


Fig,  145. 

the  different  members.  How  will  the  stresses  be  affected  by  an  addi- 
tional load  of  250  lbs.  at  each  of  the  joints  between  the  foot  and  ridge 
on  one  side  ? 

Ans.  Stress  in  BD  =5500  Vj;  DF  =  5000  \/J; 

FH  =  4500  4/5  ;  HL  —  4000  4/5";  LN=  3500  |/s"; 

NA  =  3000  4/5  ;  DE  =  o  ;  /"G  =  500 ;  HK  =  1000; 

LM=  1 500 ;  NO  —  2000  ;  AP  =  5000  ; 

BE  =  I  rooo  =  EG ;  G/v  =  loooo ;  KM  =  9000 ; 

MO  =  8000 ;  OP  =  7000 ;  DG  =  500  y'J; 

FK  =  1000  4/2  ;  HM  =  500  1/13  ;  LO  —  1000  ^J; 

NP  —  500  4/29  lbs. 


EXAMPLES. 


89 


73.  The  dead  load  upon  a  roof-truss  of  accompanying  type  consists 

of  1000  lbs.  at  F,  1000  lbs.  at  K,  and  500  lbs. 
at  G\  the  wind-pressure  is  a  normal  force  of 
30  lbs.  per  square  foot  of  roof-surface  upon 
AB;  the  span  =  90  ft.;  the  rise  =  25  ft.;  the 
trusses  are  25  ft.  centre  to  centre.  Find  the 
stresses  in  the  several  members  when  rollers 


:i229if  lbs.;  vertical  reaction 
horizontal   reaction  at  i?  = 


Fig.  146. 

are  {a)  at  C,  (b)  at  B. 

Ans. — {a)  Reaction  (vertical)  at  C- 
at   B  =  22>9SH  lbs.; 
18750  lbs. 

Tension  in  BD  =  48625  ;  DL  =  34475  ;  LE  =  21675  '< 

EC  =2212$;  DN=7S6i^;  AL  =  15888^; 

/^E  =  250  lbs. 
Compression  in  BF=^666i  y'loe;  F>Y=2788|  |/io6; 

//A  =  1977I  |/io6  ;  AA!'  =  2325  4/106  ; 

KG  =  24o8J^  4/106 ;  GC  =  2458^  4/106  ; 

DF  =  1572I  4/106;  LH-  1505^  4/181  ; 

ZA'  ^  S^i  4/T81 ;  EG=So  i^ToS  lbs. 

(6)  Only  alteration  in  stresses  is  that  each  stress  in  the 
different  sections  of  the  horizontal  tie  is  diminished 
,  by  18750  lbs.;  all  the  remaining  stresses  are  un- 

changed. 

74.  In  the  accompanying  roof-truss,  angle  ABC  =  30°  ;  the  span  =  goj^ 
ft.;  DF  =  EG  =  loj  ft.;  each  rafter  is  divided  into 
four  equal  segments  by  the  points  of  support ; 
the  trusses  are  20  ft.  centre  to  centre  ;  the  weight 
of  a  bay  of  the  roof  =  24416  lbs.     Determine  the 

Fig.  147.  stress  in  each  member. 

Also  determine  the  stresses  due  to  a  wind-pressure  of  30  lbs.  per  square 
foot  of  roof-surface  acting  normally  to  AB,  when  rollers  are  under  (a)  C, 
{f>)  B. 

75.  The  figure  represents  a  bowstring  truss  of  80  ft.  span,  cut  off  by 
the  plane  MN  and  supported  at  O.  The  upper 
flange  OCDE  is  an  arc  of  a  circle  of  85  ft.  radius ; 
OA  =  AB  =  etc.  =  10  ft.  ;  the  rise  of  the  truss 
=  10  ft. ;  a  load  of  1 5  tons  is  concentrated  at  each 
of  the  points  A  and  B\  the  reaction  at  O  =  45 
tuns.  Find  the  stresses  in  the  members  cut  by 
the  plane  MN. 


^ 


IM 


Pio.  148. 


'%^' 


m 


i 


1    .    :   ■      \\ 


.  r'-i 

..\    If 

.^;,l:„::.i^ 


r.ij. 


\      I     s 

\ 


■'-:l:i! 


■'Ml  ■' 

V    i 


]  i|l'i 


T 

53,800  lbs, 


90  THEORY  OF  STRUCTURES. 

76.  The  figure  is  a  portion  of  a  bridge-truss  cut  off  by  the  plane  MX 
and  supported  upon  the  abutment  at  A ;  AC 
=  CE-  I4,V  ft.:  the  depth  BC  =  DE  =  17^  ft. ; 
in  the  third  panel  the  compression  in  the  upper 

-<6i,6oo  lbs.  choi  1  is  64,600  lbs.;  the  tension  in  the  lower 
chord  is  53,800  lbs.  Find  the  reaction  at^^,  the 
equal  weights  supported  at  C  and  E,  and  the 
diagonal  stress  T. 

Ans.'    Reaction  =  19,474  lbs.;  weight  at  C 
and  at  £■  =  9737  lbs. ;  T=  17,977  lbs. 

77.  The  top  beam  of  a  roof  for  a  clear  span  of  96  ft.  consists  of  six 
bars  AB,  BC,  CD,  DE,  EF,  EG,  equal  in  length  and  so  placed  that 
A,  B,  C,  D,  E,  F,  G  are  on  circle  of  80  ft.  radius;  the  lower  boom  also 
consists  of  six  equal  rods  AH,  HK,  KL,  LM,  MN,  NG,  the  points  H,  K, 
L,  M,  and  N  being  on  a  circle  of  148  ft.  radius;  B  is  connected  with 
H,  C  with  K,  D  with  L,  E  with  M,  and  F  with  N \  the  opposite  corners 
of  the  bays  are  connected  by  cross-braces  ;  the  end  A  is  fixed  to  its  sup- 
port, G  being  allowed  to  slide  freely  over  a  smooth  bed-plate.  Determine 
graphically  the  stresses  in  the  various  members  when  there  is  a  normal 
wind-pressure  per  lineal  foot  of  460  lbs.  upon  AB,  340  lbs.  upon  BC,  and 
60  lbs.  upon  CD. 

78.  A  bowstring  roof-truss,  with  vertical  and  diagonal  bracing,  of 
50  ft.  rise,  and  five  panels,  is  to  be  designed  to  resist  a  wind  blowing 
horizontally  with  a  pressure  of  40  lbs.  per  square  foot.  The  depth  of  the 
truss  at  the  centre  is  10  ft.  Determine, ^;'rt'////ra//)',  the  stresses  in  the 
several  members  of  the  truss,  assuming  that  the  roof  rests  on  rollers 
at  the  windward  support. 

79.  Determine  the  chord,  vertical  and  diagonal  stresses  in  a  Howe 
truss  of  So  ft.  span,  8  ft.  depth,  and  ten  panels,  due  to  a  load  of  40  tons 
(a)  concentrated  at  the  centre  ;  (^)  concentrated  at  the  third  panel  point ; 
{c)  uniformly  distributed  ;  {d)  distributed  so  that  5  tons  is  at  first  panel 
point,  10  tons  at  second,  and  25  tons  at  third. 

Ans.  Panel  stresses  in  tension  chord  : 


I       I 


a 

1st 

2d 

3d 

4th 

5th 

6th 

7th 

8th 

9th 

loth 

20 

40 

60 

80 

IOC 

100 

80 

60 

40 

20 

h 

28 

56 

84 

72 

60 

60 

48 

36 

24 

12 

c 

18 

32 

42 

48 

50 

50 

48 

42 

32 

IS 

d 

30 

55 

70 

60 

50 

50 

40 

30 

20 

10 

EXAMPLES. 
Panel  stresses  in  compression  chord  : 


91 


a 

20 

40 

60 

80 

80 

60 

40 

20 

b 

28 

56 

84 

72 

48 

36 

24 

12 

c 

IS 

32 

42 

48 

4S 

42 

32 

18 

1 

d 

30 

55 

70 

60 

40 

30 

20 

10 

Stresses  in  verticals : 


a 

20 

20 

20 

20 

40 

20 

20 

20 

20 

b 

28 

28 

28 

12 

12 

12 

12 

12 

12 

c 

18 

14 

10 

6 

4 

6 

10 

14 

18 

d 

30 

25 

15 

ID 

10 

10 

10 

10 

10 

Diagonal  stresses : 


a 

20 

^2 

tons 

in 

each 

diago 

nal. 

b 

28V'2 

2iV2 

284/2 

124/2 

124/2 

124^2 

124^2 

124/2 

124/2 

124/2 

c 

184/2 

I4I/2 

104/2 

64/2 

24/2 

24^2 

64/^ 

104^2 

1W2 

1SV2 

d 

3oV'2 

254/2 

rsV^ 

104^2 

104^2 

104^2 

104^2 

ro4/2 

104/2 

104/2 

80.  A  Warren  girder  of  60  ft.  span,  composed  of  six  equilateral 
triangles,  carries  upon  its  lower  chord  a  weight  of  2  tons  at  the  first  and 
second  joints,  15  tons  at  the  centre  joint,  and  7^  tons  at  the  fourth  and 
fifth  joints.     Find  the  stresses  in  all  the  members. 

Am.  Stresses  in  tension  chord:  ist  bay  =  Y  4/J;    2d  =  ',"/  1^3 ; 

3d  =  s,V'  Vy^  4th  =  =^,v  Vy, 

5th  =  V  V3 ;  6th  =  i^  4' 3. 
Stresses  in  com  pr.  chord  :   ist  bay  =  '/  4/3;    2d  =  V  4/3; 

3d  =  V-  4/3J  4th  =  V  i' J: 

5th  =  V  V3. 

Diag.  stresses  ist  and  2d  bays  =  V  4/3 ;  3d  and  4th  =  -\?-  \/'2- 

5th  and  6th  =  V-  4/3  ;    7th  and    Sth  =  V-  Vy, 

9th  and  loth  =  V  V/3 ;  nth  and  12th  =  -',,"  4''3'; 

81.  Determine  the  stresses  in  the  members  of  a  Fink  truss  of  240  ft. 
span  and  sixteen  panels;  depth  of  truss  =  30  ft. ;  uniformly  distributed 
load  =  IV. 


Is;  ■'  : 

'.    t 


!i  .Mi 


:m 


h  I 


W9 


In 


liiii:- 


\  *m  '.in     J'       ■    ^  I 


, 

; 

■-It 

92 


THEORY  OF  STRUCTURES. 


-^VHD- 


Ans.—  a!   l  ft  n  o  p  q  r  s'     Stress  in  BA,  BM,  DM,  DO,  FOy  FQ, 

HQ,  HS,  same  and  — 


B   C    D    E    F    Q   H 
Fig.  150. 


64 


in 


IV 


CA,  CO,  GO,  GS  same  and  =  —7-  4/2 ; 

10 

IV     -  W    , 

m  EA,  ES  same  and  =  -5-  y'c ;  in  AK  =  —  r  17  ;   in  BL, 

o        ■'  4- 

ff  W 

DN,  FP,  HR,  same  and  =  — 7- ;  in  Ol/,  Cg  same  and  =  -5-  ; 

w  w 

in  £0  =  — -  ;  \n  KS=  —  ;  in  AM,  MO,  OQ,  OS  same  and 

64 

82.  Determine  the  stresses  in  the  members  of  a  Bollman  truss  100  ft. 
long  and  \2\  ft.  deep,  under  a  uniformly  distributed  load  of  200  tons,  to- 
gether with  a  single  load  of  10  tons  concentrated  at  25  ft.  from  one  end. 


Ans.  Stress  \n  AB  =  ^\/\;  ^Z  =  if  4/2 ;  /f Z?  =  ^  4/5  ; 

DL  =  ^- \/yj ;  AF  =  i^  i/Fo";  FL  =  ^  4/26; 

AH  -  ^  i/vj  =  HL;  in  BC  =  2$  =  FG  = 
HK  =  etc.;  DE=  50  tons;  compression  along 
/iZ  =  193I  tons. 

A''.?/^.— Questions  53,  54,  57-59,  61,  66,  67,  70,  71,  73,  and  74  can  be 
easily  solved  graphically. 

83.  Determine  the  stresses  in  the  several  members  when  the  throw 
of  the  crane  in  Question  55  is  increased  by  the  introduction  of  the  new 
members,  shown  by  the  dotted  lines. 


I'll 


rfi ' 


CHAPTER   II. 

SHEARING   FORCES  AND   BENDING  MOMENTS. 

Note. — In  this  chapter  it  is  assumed  that  all  forces  act  in  one  and  the  same 
plane,  and  that  the  deformations  are  so  small  as  to  make  no  sensible  alteration 
either  in  the  forces  or  in  their  relative  positions. 

I.  Equilibrium  of  Beams. — A  beam  is  a  bar  of  somewhat 
considerable  scantling,  supported  at  two  points  and  acted  upon 
by  forces  perpendicular  or  oblique  to  the  direction  of  its  length. 

Case  I.     AB  is  a  beam  resting  upon  two  supports  in  the 
same  horizontal  plane.     The  reactions 
A^,  and  R^  at  the  points  of  support  are 
vertical,  and   the  resultant  P  of   the 
remaining  external  forces  must  also        "^       ^? 
act  vertically  in  an  opposite  direction  ^'°'  '^'' 

at  some  point  C.     According  to  the  principle  of  the  lever, 


iA 


_c 


r^ 


^.  =  ^z^. 


R. 


P^ 
AB' 


and     R,-^R^  =  P. 


Case  II.     AB  is  a  beam  supported  ox  fixed  at  one  end. 
Such   a   support   tends   to   prevent   any  deviation    from   the 

straight  in  that  portion  of   the  beam, 
II   t  p         and   the  less   the   deviation   the  more 

^   '  1T !  ^^  perfect  is  the  fixture. 

''■q  "  ^?  The  ends  may  be  fixed  by  means  of 

^'°' '"■  two  props  (Fig.  153),  or  by  allowing  it 

^     ,3  to  rest  upon  one  prop  and  preventing 
jp       upward  motion  by  a  ledge  (Fig.  154),  or 
Fig.  154.  by  building  it  into  a  wall  (Fig.  155). 

In  any  case  it  may  be  assumed  that 


1       ' 


R 


Ac 


-[M 


I      iB  the  effect  of  the  fixture,  whether  perfect 
>ifp        or  imperfect,  is  to  develop  two  unequal 
^'°'  '55-  forces,  Q  and  R,  acting  in  opposite  di- 

rections  at  points  M  and  N,     These  two  forces  are  equivalent 

93 


\'  ■ 


'    r  ' 


(    . 


.^i  r' 


!i 


ill  I 


I    , 


rn^ 


1 


It^ 


•■i 


94 


THEORY  OF  STRUCTURES. 


to  a  left-handed  couple  (0,  —  Q),  the  moment  of  which  is 
Q.MN,  and  to  a  single  force  R—  Q  dX  N.  Hence  R  —  Q 
must  =  P. 

Case  III.    AB  is  an  inclined  beam  supported  at  A  and 

resting  upon  a  smooth  vertical  surface 
ati?. 

The  vertical  weight  P,  acting  at  the 
point  C,  is  the  resultant  load  upon  AB. 
Let  the  direction  of  P  meet  the  hori- 
zontal line  of  reaction  at  B  in  the 
point  D. 

The  beam  is  kept  in  equilibrium  by 
the  weight  P,  the  reaction  R^dX  A,  and  the  reaction  7?,  at  B. 
Now  the  two  forces  R^  and  P  meet  at  D,  so  that  the  force  R^ 
must  also  pass  through  D. 


Fic.  156. 


Hence 


R,  =  P- 


I 


cos  ADC 


and     R,  =  P  tan  ^27(7. 


A*!?/^. — The  same  principles  hold  if  the  beam  in  Cases  I  and 
n  is  inclined,  and  also  whatever  may  be  the  directions  of  the 


forces  P  and  R^  in  Case  HI, 


Case  IV.  In  general,  let  the  beam  AB  be  in  equilibrium 
under  the  action  of  any  number  of  forces  /*,,/*,,  P, ,  ...  , 
!2i  >  (2(1  >  G3 .  •  •  •  >  of  which  the  magnitudes  and  points  of  appli- 


At 


Fig.  157. 

cation  are  given,  and  which  act  at  right  angles  to  the  length  of 
the  beair.  Suppose  the  beam  to  be  divided  into  two  segments 
by  an  imaginary  plane  MN.  Since  the  whole  beam  is  in  equi- 
librium, each  of  the  segments  must  also  be  in  equilibrium. 
Consider  the  segment  AMN. 


?  ^^  ^'^ 

?' 

f 

1 

N 

i    1 

V       V 
Qa   Q2 

EQUILIBRIUM  OF  BEAMS. 


95 


It  is  kept  in  equilibrium  by  the  forces  P^,  P^,  P^,  .  .  .  and 
by  the  reaction  of  the  segment  BAIN  upon  the  segment  AMN 
at  the  plane  MN\  call  this  reaction  E^.  The  forces/*,,/*,, 
I\,  .  .  .  are  equivalent  to  a  single  resultant  R^  acting  at  a  point 
distant  r,  from  AIN.  Also,  without  affecting  the  equilibrium, 
two  forces,  each  equal  and  parallel  to  /v, ,  but  opposite  to  one 
another  in  direction,  may  be  applied  to  the  segment  AMN  at 
the  plane  MN,  and  the  three  equal  forces  are  then  equivalent 
to  a  single  force  R^  at  MN,  and  a  couple  {R^ ,  —  R^)  of  which 
the  moment  is  /?,r, . 


Ru  ^jT'^-^Ru^R.'^'^ 


33 


I 


V_-ii^ 


Fig.  158. 


Thus  the  external  forces  upon  AMN  are  reducible  to  a 
sincflo  force  /?,  at  AIN,  and  a  couple  {R^ ,  —  R^).  These  must 
bo  balanced  by  E^ ,  and  therefore  E^  is  equivalent  to  a  single 
force  —  R,  at  A/N  and  a  couple  {—  R,,  /?,). 

In  the  same  manner  the  external  forces  upon  the  segment 
BMN  are  reducible  to  a  single  force  /?,  at  AIN,  and  a  couple 
(A\,  —  R^  of  which  the  moment  is  R^r^.  These  again  must 
be  balanced  by  E^,  the  reaction  of  the  segment  AMN  upon 
the  segment  BMN. 

Now  /",  and  E^  evidently  neutralize  each  other,  so  that  the 
force  /?,  and  the  couple  (/l,  ,  —  /?,)  must  neutralize  the  force 
A\  and  the  couple  {R^,  —-^0-  Hence  the  force /?,  and  the 
couple  {R^ ,  —  R,)  are  respectively  equal  but  opposite  in  effect 
to  the  force  R,  and  the  couple  {R^,  —  R,) ;  i.e., 


.■'§1' J 


\\ 


<  r< 


7?,  =  Rj    and     Rr^  ^  R^r^ ; 


r,  =  r,. 


The  force  R^  tends  to  make  the  segment  AMN  slide  over 
the  segment  BMN  aX.  the  plane  AIN,  and  is  called  the  Shearing 


53  .  I- 


I  ft 


^    J     1 


II 


^il 


96 


THEORY  OF  STRUCTURES. 


Force  with  respect  to  that  plane.     It  is  equal  to  the  algebraic 
sum  of  the  forces  on  the  left  of  MN, 

=  p^  +  p^-p^j^...  =  :s{P). 

So  ^,  =  (2,  —  ^3  —  S3  +  •  •  =  ^(0  is  the  algebraic  sum 
of  the  forces  on  the  right  of  MN,  and  is  the  force  which  tends 
to  make  the  segment  BAIN  slide  over  the  segment  AMN  at 
the  plane  J/A''.  ^3  is  therefore  the  Shearing  Force  -wXih.  respect 
to  MN,  and  is  equal  to  R^  in  magnitude,  but  acts  in  an  opposite 
direction. 

Again,  let/, ,  A .  A »  •  •  •  .  ^1 »  ^» .  ^3 .  •  •  •  .  be  respectively  the 
distances  of  the  points  of  application  of  /*,,/!,,  P, ,..  .  ^Q^,Q^, 
<2, ,  .  .  .  from  MN. 

Then  R^r^ ,  =  the  algebraic  sum  of  the  moments  about 
MN  of  all  the  forces  on  the  /eft  of  AfN, 

=  P.P.  +  P.P.  -  /'3 A  +  •  .  .  =  ^{Pp\ 

is  the  moment  of  the  couple  iR^,  —  /?,). 

This  couple  tends  to  bend  the  beam  at  the  plane  MN,  and 
its  moment  is  called  the  Bending  Moment  with  respect  to  MN 
of  all  the  forces  on  the  left  of  MN. 

So  /v',;-, ,  :=  the  algebraic  sum  of  the  moments  about  il/A^ 
of  all  the  forces  on  the  right  of  MN, 


=  Q.Q.  -  Q.^. 


=  ^{QqI 


is  the  Bending  Moment,  with  respect  to  MN,  of  all  the  forces  on 
the  right  of  MN,  and  is  equal  but  opposite  in  effect  to  R^r^ . 

It  is  seen  that  the  Shearing  Force  and  Bending  Moment 
change  sign  on  passing  from  one  side  of  MN  to  the  other,  so 
that  to  define  them  absolutely  it  is  necessary  to  specify  the  seg- 
ment under  consideration. 

Remark. — The  reaction  E^  has  been  shown  to  be  equivalent 
to  the  force  —  R^  and  the  couple  (—  R^ ,  /?,).  The  Moment 
of  this  couple  may  be  called  the  Elastic  Moment,  the  Moment 
of  Resistance,  or  the  Moment  of  Inflexibility,  and  is  equal  in 
magnitude,  but  opposite  in  effect,  to  the  corresponding  Bend- 
ing Moment  due  to  the  external  forces. 


2.  E: 

ments.- 
example 
of  length 

Ex. 
is  fixed 
/'at  O. 

The  j 
point  of 
stant  and 
Upon 
equal  or 
between  ; 
shearing  i 
Again 
distant  x 
Upon 
tional   to 
point  of 
moment  a 
Ex.  2. 


take  AB  e( 
distance  be 
resents  the 

Again,  t 
is  nil  at  (9, 
^C  equal  o 


SHEARING  FORCES  AND  BENDING  MOMENTS. 


97 


Fig.  139. 


2.  Examples  of  Shearing  Forces  and  Bending  Mo- 
ments.— In   each   of   the  following 
examples  the  beam  is  horizontal  and 
of  length  /. 

Ex.  I.  The  beam  OA,  Fig.  159, 
is  fixed  at  A  and  carries  a  weight 
/'at  O. 

The  Shearing  Force  {S)  at  every 
point  of  the  beam  is  evidently  con- 
stant and  equal  to  P. 

Upon  the  verticals  through  A  and  O  take  AB  and  (9Ceach 
equal  or  proportional  to  P\  join  BC.  The  vertical  distance 
between  any  point  of  the  beam  and  the  line  BC  represents  the 
shearing  force  at  that  point. 

Again,  the  Bendbig  Moment  (M)  at  any  point  of  the  beam 
distant  x  from  O  is  Px ;  it  is  nil  at  O,  and  P/  at  A. 

Upon  the  vertical  through  A  take  AD  equal  or  propor- 
tional to  P/;  join  DO.  The  vertical  distance  between  any 
point  of  the  beam  and  the  line  DO  represents  the  bending 
moment  at  that  point. 

Ex.  2.     The  beam  OA,  Fig.  160,  is  fixed  at  A,  and  carries 

a  uniformly  distributed  load,  of  in- 
tensity zu  per  unit  of  length. 

The  resultant  force  on  the  right 
of  a  vertical  plane  J/iVdistant  x  from 
O  is  wx  and  acts  half-way  between 
O  and  MNi 

The  SJiearing  Force  (S)  at  MN  is 
therefore  zvx ;  it  is  nil  at  O,  and  ivl 
at  A.  Upon  the  vertical  through  A 
take  AB  equal  or  proportional  to  wl\  join  BO.  The  vertical 
distance  between  any  point  of  the  beam  and  the  line  BO  rep- 
resents the  shearing  force  at  that  point. 

X       tax' 
Again,  the  Bending  Moment  (M)  at  A/N  is  tux-  — ;  it 

is  nil  at  O,  and  — -at  A.     Upon  the  vertical  through  A  take 

zvi' 
J  C  equal  or  proportional  to . 


^w.x 


Fig.  160. 


■^ 


mu\ 


U:l*Ll 


THEORY  OF  STRUCTURES. 


I  s 


■: 


The  bending  moment  at  any  point  of  the  beam  is  repre- 
sented by  the  vertical  distance  between  that  point  and  a  pa- 
rabola CO  having  its  vertex  at  0  and  its  axis  vertical. 

Ex.  3.     The  beam  OA,  Fig.  161,  is  fixed  <i.\.  A  and  carries 


Fig.  161. 

a  single  weight  P  at  O,  together  with  a  uniformly  distributed 
load  of  intensity  xv  per  unit  of  length. 

The  Shearing  Force  {S)  at  a  plane  MN  distant  x  from  0 
is  evidently  P  -\-  %vx  ;  it  is  P  at  0,  and  P -\-  wl  zX  A. 

Upon  the  verticals  through  0  and  A  take  OC  equal  or  pro- 
portional to  P,  and  AB  equal  or  proportional  to  ivl-\-P\ 
join  BC.  The  vertical  distance  between  any  point  of  the 
beam  and  the  line  BC  represents  the  shearing  force  at  that 
point. 

Again,  the  Bending  Moment  {jlf)  at  AIJV  is  evidently 


X  7('X 

zvx — |-  Px  = (-  Px ; 


It  is  nil  at  0,  and h  PI  at  A. 

Upon   the  vertical  through  A  take  AD  equal  or  propor- 

tional  to 1-  PL     The  bending  moment  at  any  point  of 

the  beam  is  represented  by  the  vertical  distance  between  that 
point  and  a  parabola  DOE  having  its  axis  EF  vertical  and  its 


vertex  a 

portiona 

Note. 
the  algcb 
lines  BC 
curve  DL 
rcspondii 
CO  in  E 
arrived  ai 
pcndentl} 

Ex.  4 
at  0  and 
at  a  poi 
into  the  t 
which  th( 
spectivel) 

The 
A  are  ver 
principle  c 


The  S) 

constant  a 
fiircc  {S)  c 
ami  A  is  co 
I  he  vertica 
l)roportion 


to 


Pa 


the  beam  ii 
point  and  t 


Jl.. 


SHEARING  FORCES  AND  BENDING  MOMENTS. 


99 


vertex  at  a  point  E,  where  OF  =  -  and  EF  is  equal  or  pro- 

w 


poitional  to 


2tv 


Note. — The  ordinates  of  the  line  BC  in  Ex.  3,  are  equal  to 
the  algebraic  sum  of  the  corresponding  ordinates  of  the  straight 
lines  BC  and  BO  in  Exs,  i  and  2.  Also,  the  ordinates  of  the 
curve  DO  in  Ex.  3,  are  equal  to  the  algebraic  sum  of  the  cor- 
responding ordinates  of  the  line  DO  in  Ex.  i,  and  the  curve 
CO  in  Ex.  2.  Hence  the  same  conclusions  as  in  Ex.  3  are 
arrived  at  by  treating  the  weight  P  and  the  load  ivl  inde- 
pendently, and  then  superposing  the  respective  results. 

Ex.  4.  The  beam  OA,  Fig.  162,  rests  upon  two  supports 
at  0  and  A,  and  carries  a  weight  P 
at  a  point  B,  dividing  the  beam 
into  the  two  segments  OB,  BA,  of 
which  the  lengths  are  a  and  b  re- 
spectively. 

The  reactions  ^1 ,  R^  at  C?  and 
A  are  vertical,  and  according  to  the 
principle  of  the  lever,  Fig.  163. 

R,  =  P-^     and     R,  =  Pj 


G 

c 

/ 

E      \ 

t"' 

/ 

\, 

0 

/ 

B 

\ 

A 

'■^''^ 

. 

F 

D 

■  -.'-^ 

" 

3 

V 

\      \ 


\  %% 


;'i        I J 


The  Shearing  Force  (5)  at  every  point  between  O  and  B  is 
constant  and  equal  to  /?,  =  P  ..  On  passing  B  the  shearing 
force  (5)  changes  sign,  and  its  value  at  every  point  between  B 


and  A  is  constant  and  equal  to  R^  —  P 


r']  =  -  A'. 


Upon 


the  verticals  through  (9,  Z>',  and  A  take  OC,  BE,  each  equal  or 

Pb 
[)roportional  to  -v-,  and  BF,  AD,  each  equal  or  proportional 

to  -  y  ;  join  CE  and  DF.     The  shearing  force  at  any  point  of 

the  beam  is  represented  by  the  vertical  distance  between  that 
point  and  the  broken  line  CEFD. 


t  'is 


•<4-! 


WMil 


1 1  i 


ri     i: 


li'l  1  » 


lilMi 


F     ~ 


iJiii-  '■  f. 


lOO 


THEORY  OF  STRUCTURES. 


It      IS 


Again,  the  Bending  Moment  (M)  at  any  point  between  O 
and  ^  distant  x  from  6>  is  R^x  =  P-.x\  it  is  nil  at  O,  and 

P^*  at  /i. 

The  Bending  ^lament  (Af)  at  any  point  between  B  and  A 
distant    x    from    6>    is   R,x  —  /'(;«;  —  a)  =  P-A/  —  x) 

P-.  at  B,  and  nil  at  y?. 

Upon  the  vertical  through  B  take  BG  equal  or  proportional 
to  P—f\   join   6^6^   and  AG.     The   bending   moment    at   any 

point  of  the  beam  is  represented  by  the  vertical  distance  be 

twecn  that  point  and  the  line  OGA. 

P 
tor. — If /*bc  at  the  centre  of  the  beam,  5  =  -,   and  M 


at  the  centre  = 


PI 


Ex.  5.     The  beam  OA,  Fig.  163,  rests  upon  two  supports 
p  at    C?   and   A,  and    carries   a  uni- 

formly  distributed  load  of  inten- 
sity zv  per  unit  of  length. 

The  reactions  at  0  and  A  are 
wl 


each  equal  to  — 

The  resultant  force  between  0 

and  a  plane  MN  distant  x  from  0 

Fig.  163-  is  wx,  and  acts  half-way  between 

0  and  MN.     The    Shearing  Force  (S)  at  MN  is   theiefoie 

wl  wl 

zvx;    it    is  at   O,   nil   at   the   middle   point   B,  and 

at  A.     Upon  the  verticals  through  O  and  A  take  OC 

wl 
and  ^Z?,  each  equal  or  proportional  to  - — ;  join  CJy.     The 

shearing  force  at  any  point  of  the  beam  is  represented  by  the 
vertical  distance  between  that  point  and  the  line  CD. 


SHEA  KING  FOJtCES  AND  BENDING  MOMENTS.         lOI 


Again,  the  Bending  Moment  (Af)  at  MN  is 


w/             X      ti'i         Tta' 
— X  —  wx-  =  — X 

2.22  2 


it  is  nil  at  O  and  ^t  A  \  it  is  a  maximum  and  equal  to  --—  at 

o 

the  middle  point  B.  Upon  the  vertical  through  B  take  BE 
equal  or  proportional  to  -  -.     The  bending  moment  at  any 

point  of  the  beam  is  represented  b\'  the  vertical  distance  be- 
tween that  point  and  a  parabola  OEA  having  its  vertex  at  £ 
and  its  axis  vertical. 

Cor.  I.  The  shearing  force  is  a  minimum  and  zero  at  the 

centre,  a.  maximum  and  —  at  the  ends,  and  increases  uni- 

2 

formly  with  the  distance  from  the  centre. 

Cor.  2.  The  bending  moment  is  a  minimum  and   zero  at 


the  ends,  a  maximum  and 


■wT 
8 


at  the  centre,  and  diminishes 


as  the  distance  from  the  centre  increases. 

Ex.  6.  The  beam  OA,  Fig.  164,  rests  upon  two  supports  at 
0  and  A,  and  carries  a  weight  P  at 
a  point  B,  together  with  a  uniformly 
distributed  load  of  intensity  u>  per 
unit  of  length. 

Let  the  lengths  of  the  segments 
OB,  BA  be  a  and  b,  respectively. 

The  reactions  /?,  at  0,  and  R^ 
at  A,  are  vertical,  and  according  to 
the  principle  of  the  lever, 


R. 


and 


Fig.  164. 


'<!t!..i'9 


•i!Hii! 


t  )« 


m 


\.\\ 


H 


1'  I 


V^\l\ 


C'l 


m 


i.-^ij 


! 


102 


THEORY  OF  STRUCTURES. 


The  Shearing  Force  {S)  at  any  vertical  plane  between  O  and 
B  distant  x  from  O  is 

b      wl 

/?,  —  wx  =  P-.  -| wx ; 

.    ,      Ph       ivl        ^  ,     Pb    ,   w/ 

It  IS    -/-  H at  O,    and     --r-  -] «/«  at  5. 

The  Shearing  Force  {S)  at  any  plane  between  ^  and  ^i 
distant  x  from  (7  is 


.^b  .   W 


w/ 


<? 


1  -  ■  1 

; 

jl 

1  ■ 

:   ■ 

r 

m 

\m 

^iii;  i  , 

vm 

:;  f  -:;   ■ 

li,  —  P  —  wx  =  P~  -\ —  P—  wx  =  -^  —  P".  —  lux  ; 

'  /   '     2  2  / 


.    .      tvl      Pa  _  , 

it  IS       —. zva  at  B,    and 

2  > 


Prt        wl 

-, at  ^. 

/  2 


Upon  the  verticals  through  O,  B,  and  y4   take  OC  equal  or 

.       ,          Pb    ,    wl      „^ 
proportional    to    —,-  -\ ,    BD    equal    or    proportional    to 

Pb    ,   7vl  „^  ,  .       ^        ivl       Pa 

_| ^(i^  }jj^  equal  or  proportional  to ^ wa, 

wl       Pa 
and  /i/^  equal  or  proportional  to j  ;  join  CD  and 

EF.  The  shearing  force  at  any  point  of  the  beam  is  rep- 
resented by  the  vertical  distance  between  that  point  and  the 
broken  line  CDEF. 

wl       Pa 
\i  —  >  ~r  +  ^^^.  BE  is  positiv^e,  and  therefore  E  is  ver 

tically  above  B. 

Again,  the  Bending  Moment  (M)  at   any  point  between 
O  and  B  is 

b      wl\        wx^ 


H+x 


;^ 


it  is  nil  at  O,  and 


( ^b  ,   wl\         wa* 
\p-^--]a-~^,B. 


and 


SHEARING  FORCES  AND   BENDING  MOMENTS. 


103 


The  bending  moment  (J/)  at  any  point  between  B  and  A 
distant  x  from  O  is 

(b   ,    tvl\         zva'        r^        ,     .,        . 
It  IS  \P-,  -\ \a at  ^,  and  ml  at  A. 

Upon  the  vertical  through  B  take  BG  equal  or  propor- 

/    b       tul\         wa^ 
tional    to    \P-.-\---\a .      The    bending    moment    at 

any  point  of  the  beam  between  O  and  B  is  represented  by  the 
vertical  distance  between  that  point  and  a  parabola  OGH 
having  its  axis  HT  vertical  and  its  vertex  at  a  point  //,  where 


and 


0T=  --LP-7  +  - 


•  I  /  ^    wiy 

HT  is  equal  or  proportional  to  -—[P-i  +  — J  • 


The  bending  moment  at  any  point  between  B  and  A  is 
represented  by  the  vertical  distance  between  that  point  and  a 
parabola  AGK  having  its  axis  A'F  vertical  and  its  vertex  at  a 
point  A",  where 


w\  /    '      2  /' 


and 


KV  is  equal  or  proportional  to 


2ZV 


Pi  +  tU""- 


Cor. — If  the  weight  /*  is  at  the  centre, 


P  PI         UfP 

S  =  — ,  and  M  at  the  centre  = 1 — ^r-. 

2  '  48 

Note. — The  ordinates  of  the  lines  CD  and  EF  in  Ex.  6  are 
equal  to  the  algebraic  sum  of  the  corresponding  ordinates  of  the 


.!  'i 


n 


8..  .-iii-l-t 


i* 


104 


THEORY  OF  STRUCTURES. 


' 


lines  CE,  FD  in  Ex.  4,  and  the  line  CD  in  Ex.  5.  Also,  the 
ordinates  of  the  curves  OG,  AG  are  equal  to  the  algebraic 
sum  of  the  corresponding  ordinates  of  the  lines  OG,  AG  in  Ex. 
4,  and  the  curve  OEA  in  Ex.  5.  Hence  the  same  conclusions 
as  in  Ex.  6  are  arrived  at  by  treating  the  weight  P  and  the 
load  ivl  independently,  and  then  superposing  the  respective 
results. 

Ex.  7,  In  fine,  a  beam,  however  loaded,  may  be  similarly 
treated,  remembering  that  if  the  load  changes  abruptly  at  dif- 
ferent points,  the  portions  of  the  beam  between  the  points  of 
discontinuity  are  to  be  dealt  with  separately.  For  example, 
the  beam  OA,  Fig.  165,  rests  upon  two  supports  at  0  and  A, 
and  carries  three  weights  /*, ,  P^,  P^  at  points  C,  D,  E,  of  which 
the  distances  from  O  are  />,,  />, ,  p^,  respectively.  A  point  B 
divides  OA  into  segments  OB  =  a  and  BA  =  b,  which  are 


i  I 


F 

r^G 

/^ 

^■^T?"^--^^^ 

\ 

/     IC 

1 
1 

Mk 

\ 

0 

l« 

iD^"~^v. 

E  \ 

A 

'?^' 

f~ 

i 

y 
Pi 

P« 

N 

1 
1 

N 

K 

^. 

P3  ^ 

T 

Fig    165. 

uniformly  loaded  with  weights  of  intensities  w,  and  w,  per 
unit  of  length,  respectively.  The  reactions  7?,  and  R^  at  O  and 
A  are  vertical,  and  according  to  the  principle  of  the  lever, 

R,l  =  />.(/  -  /.)  +  Pil  -  A)  +  Pit  -  A) 


and 


Kl  =  P. P.  +  P.P.  +  P^^  +  ^-  +  tvb[-  4-  4 


SHEARING  FORCES  AND  BENDING  MOMENTS. 


105 


To  represent  graphically  the  Shearing  Force  at  different 
points  of  the  beam  : 

Upon  the  verticals  through  O,  C,  B,  D,E,  A,  take  OF,  CG, 
CH,  BK,  DL,  DM,  EN,  EV,  and  ^r,  respectively  equal  or 
proportional  to 


^f^ltHI 


^. 

R,  —  w,p„  R  -  7v,p,  -  P,,R,~  7i\a 

R. 

—  zc\a  —  P,  —  7i',{f,  —  a), 

R. 

-  Zi\a,  —  P,  —  w,(A  -  (i)-  P.„    . 

A'. 

-  w.a  —  P,  —  7i\ip,  —  (I)  -  P„ 

A', 

-  n      -P,-  xi'lp,  -a)-P,-  P„ 

A, 

-  ZV,..      -  P,  -  ZVJ)  -  P,-P,:=  R,. 

P., 


ind 


Join  FG,  HK,  KL,  MN,  and  VT.  The  shearing  force  at 
any  point  of  the  beam  is  represented  by  the  vertical  distance 
between  that  point  and  the  broken  line  FGHKLMNVT. 

To  represent  graphically  the  Bending  Moment  {M)  at  dif- 
ferent points  of  the  beam  : 


Mat  0  =  0;        Mat  C=  Rj>,  — 


w^p^ 


7U  rt;' 
MatB-  R,a '- P,{a  -  p,) ; 


M 


atD  =  R,p,  -  w,a[p,  -  I) 


--,^^'-/'.(A-A). 


.'"i 


ti 


t  h 
\  1 


1'  I 


1. 


I  *  i  I 


I^J 


io6 


THEORY  OF  STRUCTURES. 


MB.tE  =  R,p,  -  2va[p,  -  I) 


—  w. 


^.0>,-A)-^.(A-A); 


and 


Mat  A=  o. 


Upon  the  verticals  through  C,  B,  D,  and  E  take  Ci,  ^2, 
Z?3,  and  £'4,  respectiv^-ly  equal  or  proportional  to  the  bending 
moments  at  these  points. 

The  bending  moment  at  any  point  of  the  beam  is  repre- 
sented by  the  vertical  distance  between  that  point  and  the 
parabolic  arcs  Oi,  12,  23,  34,  and  4A.  The  axes  of  these  pa- 
rabolas are  vertical,  and  the  positions  of  the  vertices  may  be 
easily  found  from  the  several  equations. 

Ex.  8.  A  beam  OA,  Fig.  166,  of  which  the  weight  may  be 

neglected,  is  15  ft.  long,  is  fixed 
at  O,  and  carries  a  weight  of  80 
lbs.  at  A.     Determine  the  bend- 
ing moment  at  a  point  distant  10 
ft.  from  the  free  end.     Also  illus- 
trate the  shearing  force  and  bend 
ing  moment  at  different  points  of 
the   beam   graphically.      The   re- 
quired bending  moment  is  80  X 
10  =  800  Ib.-ft. 
The  shearing  force  is  the  same  at  every  point  of  the  beam, 
and  eq-ial  to  80  lbs.     Choose  a  vertical  scale  of  measurement 
so  that  half  an  inch  represents  160  lbs. 

Upon  OA  describe  a  rectangle  OABC,  in  which  OC  =  AB 
=  ^".  The  ordinate  from  every  point  of  BC  to  AO  is  i",  or 
80  lbs.,  and  is  therefore  the  shearing  force  at  the  foot  of  such 
ordinate. 

Again,  the  bending  moment  at  O  is  80  X  15  =  1200  Ib.-ft. 
Choose  a  vertical  scale  of  measurement  so  that  i  inch  repre- 
sents 1200  Ib.-ft.     Upon  the  vertical  through  O  take  OD  ^ 


Fig.  166. 


SHEARING  FORCES  AND   BENDING  MOMENTS. 


\0J 


I  inch ;  join  DA.  The  ordinate  from  any  point  of  DA  to  OA 
is  the  bending  moment  at  its  foot.  For  example,  at  \\\  ft. 
from  0  the  ordinate  is  i",  or  300  Ib.-ft.,  and  this  is  equal  to 
80  X  3$',  i-e.,  the  bending  moment. 

Ex.  9.  A  beam  OA,  Fig.  167,  of  which  the  weight  may  be 
neglected,  rests  upon  tw^  supports 
at  0  and  A,  30  ft.  apart,  and  carries 
a  uniformly  distributed  load  of  200 
lbs.  per  lineal  foot,  together  with  a 
single  weight  of  600  lbs.  at  a  point 
B  dividing  the  beam  into  segments 
OB,  BA,  of  which  the  lengths  are  10 
and  20  ft.  respectively.  Determine 
the  shearing  force  and  bending  mo- 
ment at  the  points  C  and  D,  distant 
5  ft.  from  the  nearest  end.  Also, 
illustrate  graphically  the  shearing 
force  and  bending  moment  at  differ- 
ent points  of  the  beam. 

Let  R^ ,  R^  be  the  reactions  at  O 
and  A,  respectively.     Then 

.^, .  30  =  600  .  20  +  200 .  30 .  1 5  =  102000 ; 

.•.  R,  =  3400  lbs.,  and  R^  =  200  .  30  +  600  —  /?,  =  3200  lbs. 

The  Shearing  Force  ?Lt  C  =  3400  —  200  .  5  =  2400  lbs. 
"  "  "       "  Z>=  3400— 200.  25— 600=  — 2200  lbs. 

The  Bending  Moment  at  C  =  3400 .  5  —  200  .  5  .  - 

=  14,500  Ib.-ft. 

25 
The  Bending  Moment  at  Z)  =  3400. 25  —  200 .  25 .  —  —  600. 1 5 

=  13,500  Ib.-ft. 

Next,  considering  the  segment  OB,  the  shearing  force  at  O 
is  3400  lbs.,  and  at  B  1400  lbs. 

Considering  the  segment  BA,  the  shearing  force  at  A  is 
—  3200  lbs.,  and  at  B  800  lbs. 

Choose  a  vertical  scale  of  measurement  so  that  i  inch  repre- 
sents 3000  lbs.  Upon  the  verticals  through  0,  B,  A  take  OE 
=  lyV'.  ^^=  iV.  -^^  =  tV'»  and  AH=.  i^-g"  ;  join  £Fand 


Fig.  167. 


'   1'     i^ 

1         '! 


i     1 


I  '         1 


1     1, 


I         I  I 


S  iP, 


io8 


THEORY  OF  STRUCTURES. 


GH.  The  ordinate  from  any  point  of  the  broken  line  EFGH  to 
OA  is  the  Shearing  Force  at  its  foot.  For  example,  the  ordinate 
at  Z>  is  —  \\" ,  or  —  2200  lbs. 

Again,  the  bending  moment  at  B  is  3400.  10—  200.  10. 5 
=  24,ocx3  Ib.-ft.  Choose  a  vertical  scale  of  measurement  so  that 
I  inch  represents  24,000  Ib.-ft.  Upon  the  vertical  through  />' 
take  BK=^  I  inch.  Draw  the  parabolas  OK,  AK,  with  their 
vertices  at  points  determined  as  in  Example  (6).  The  ordi- 
nate from  any  point  of  the  curves  OK,  AK  is  the  bending 
moment  at  its  foot. 

For  example,  at  a  point  14  ft.  from  O  the  curve  ordinate  is 
1^1^"^  or  25,600  Ib.-ft.,  and  this  is  the  Bending  Moment  at  the 
same  point,  being  also  the  greatest  for  the  segment  BA.  The 
vertex  of  AK  is,  therefore,  vertically  above  the  point  of 
which  the  horizontal  distance  from  6>  is  14  ft. 

3.  Relation  between  Shearing  Force  and  Bending 
Moment. — Let  a  beam  AB  be  arbitrarily  loaded  with  weights 
w, ,  zo^,  w^,  .  .  .  concentrated  at  the  points  i,  2,  3,  .  .  . 


The^ 

i/,,  at  A 
M,  "    I  = 
J/,  "   2  = 

i/„  "   n  = 


Hence 
ginning  an 
shearing  fi 
tcrval. 

Let  J  J 
bending  m 


r~i 


r-H 

I  .11 


,  Fig.  168. 

Let  a,,  a,,  a^,  .  .  .he  the  lengths  of  the  segments  Ai,  12, 
23,  ...  ,  respectively. 

Let  M^i ,  Mb  be  the  moments  at  A  and  B.  These  moments 
are  of  course  m/  if  the  beam  merely  rests  upon  supports  at  its 
ends. 

The  reaction  R  at  A  is  given  by  the  equation 

Rl  =  wil  -  a,)  +  wll  -a,-a,)-\-...-^MB-\-M^, 
1  being  the  length  of  the  beam. 

The  shearing  force  S,  between  A  and  i  =  7? ; 

5,        "         I     "     2  =  ^  -  7f , ; 

.S",        "         2     "     3  =  /?  -  zv,  -  w, ; 


This  re 

the  nuntbc? 

number  an 

Thus,  i: 


or,  the  shea 
of  the  bcndi 

The  abc 
shearing  fo) 
at  the  corr 
curve. 

The  she 


5„        "     n-i  "     n  =  R-'2{w)\ 
^(w)  denoting  die  sum  of  the  first  («  —  i)  weights. 


.^ 


Li-; 


SHEARING    ^ORCE  AND  BENDING  MOMENT.  IO9 

The  bending  moment 

Mj  at  A=M^\ 

J/,  "   2  =  R{a,  +  a,)  -  w,a,  +  J/,  =  i^,  +  S,^;, ; 

Hence  ///f"  difference  betzveen  the  bending  moments  at  the  be- 
ginning and  end  of  any  interval  is  equal  to  the  product  of  the 
shearing  force  (5)  for  that  interval  by  the  length  (a)  of  the  in- 
terval. 

Let  AM  denote  the  difference  between  any  two  consecutive 
bending  monnents  ;  then 

Mf  =  Sa. 

This  result  has  been  deduced  without  any  assumption  as  to 
the  number  of  the  loads.  They  may  therefore  be  infinite  in 
number  and  in  the  limit  form  a  continuous  load. 

Thus,  if  5  be  the  shearing  force  at  a  distance  x  from  /3, 


d_M_ 
dx 


=  3;  ' 


or,  the  shearing  force  at  any  point  is  eqtial  to  the  rate  of  increase 
of  the  bending  moment  per  unit  of  length. 

The  above  results  may  also  be  expressed  as  follows :  The 
shearing  force  at  any  point  is  measured  by  the  tangent  of  the  slope 
at  the  corresponding  point  of  the  b ending-moment  polygon  or 
curve. 

The  shearing  force  is  positive^  zero,  or  negative  according  as 


R  =  ^{w) ; 
< 


wT't 


^■\ 


» 'J  ilt  fiix^. 


i 
1 


:.iS' 


•■.  .;,i. 


1     < 


i ) 


V 


\  'if 


\\ 


'  \ 


I 


I  mi 


■I       11 

■t/  HI 


'\\\V 


i 
111  I 

1      '' 

1  ■■ 


no 


THEORY  OF  STRUCTURES. 


^(w)  being  the  sum  of  the  weights  up  to  the  point  under  con- 
sideration.    In  the  case  of  a  continuous  load,  of  intensity  cb, 


f  H  =  f 


wdx. 


Thus  the  bending  moment  iJ/at  the  same  point  is  a  maxi- 
mutn  (or  a  minimum  in  certain  special  cases)  when  the  shear- 
ing force  changes  sign,  i.e.,  when 

S  —  o. 
Again,  with  an  arbitrarily  distributed  load 

and  with  a  continuous  load 


Thus  the  difference  between  the  ordinates  of  the  bending- 
moment  diagram  at  any  point  and  A  is  proportional  to  the 
area  of  the  shearing-force  diagram  between  the  same  points. 
From  this  result  an  important  deduction  may  at  once  be 
made. 

The  bending  moment  M^  at  any  point  between  r  and  r  +  i 
distant  x  from  r  is 

M.,  =  R{a,-\-a^-\-  . . .  -\-a,-\-x)—wSa^-\-a^-{- . . .  -{•ar-\-x)-\- . . . 

-WrX-\-M^ 

=  My  -\-  x{R  —  zv^  —  w^  —  .  .  .  ~  Wr) 

Now    My^.^  =  My  -\-  «!,.+i5,.+i,         and  therefore  5,.+,  is  zero  if 
Mr^^  =  My,    and  also    M^  =  My  =  My^^. 

Thus,  the  bending  moment  is  the  same  at  every  point  between 
r  and  r  +  i .  (i^^d  the  case  is  one  of  simple  bending  without  shear, 
as,  e.g.,  with  a  carriage-axle. 


4.  T( 

Let  a  sin 
OA   of   1 
supports 
The 

■  at  B  dist 

and  is  tl 
points  be 
IV  accorc 
or  0.  U 
take  OD  ( 
force  at  a; 
the  latter 
tical  dista 
Also,  t 

R,  -  W  -- 

distance  b 

to  OD. 

Again, 


Fi 

tween  the  { 
and,  its  vert 

Note.—", 
on  both  sid 
with  one  ha 

Cor.  I. 
point  are  m, 

For  exai 


i 


I 


EFFECT  OF  A    ROLLING   LOAD. 


Ill 


4.  To  Discuss  the  Effect  of  a  Rolling  Load. — Case  I. 
Let  a  single  wciglit  W  travel  from  left  to  right  over  a  girder 
OA   of   length  /,  resting    upon  two 
supports  at  O  and  A, 

The  reaction  /?,  at  O,  when  W  is 

at  B  distant  x  from  0,  is   W — -. — 


Fig.  169. 


and  is  the  Shearing  Force  for  all 
points  between  O  and  B\  it  is  nil  or 
\V  according  as  the  weight  is  at  A 
or  0.  Upon  the  vertical  through  O 
take  OD  equal  or  proportional  to  W\  join  DA.  The  shearing 
force  at  any  point  of  tlie  beam  between  O  and  the  weight,  as 
the  latter  travels  from  A  towards  O,  is  represented  by  the  ver- 
tical distance  between  that  point  and  the  line  AD. 

Also,  the  shearing  force  at  any  point  between  B  and  A  is 

X 

R^—  W  =  —  W-.,  and  is  equal  or  proportional  to  the  vertical 

distance  between  that  point  and  the  line  OE  where  AE  is  equal 
to  OD. 

Again,    the   Bending  Moment   at   B,  when   W  is   at  B,  is 

—X ;  It  IS  ml  at  O  and  at  A  : 


W- 


it  is  a  maximum  and  = 


Wl 


at  the 


Fig.  170. 


middle  point  D.  The  bending  mo- 
ment at  any  point  of  the  beam  when 
the  weight  is  at  that  point  is  repre- 
sented by  the  vertical  distance  be- 
tween the  point  and  the  parabola  OEA,  having  its  axis  vertical 

]Vi 
and,  its  vertex  at  E,  where  DE  is  equal  or  proportional  to . 

Note. — The  shearing  and  bending  actions  are  symmetrical 
on  both  sides  of  the  centre,  and  it  is  therefore  sufficient  to  deal 
with  one  half  of  the  girder  only. 

Cor.  I.  The  shearing  force  and  bending  moment  at  any 
point  are  maxima  at  the  instant  the  weight  passes  that  point. 

For  example,  the  shearing  force  at  B  for  the  segment  OB, 


h « ' 


'       Mi 


fii 


1    it!  ■ 


II  iw 


112 


THEORY  OF  STRUCTURES. 


i      \ 


%  I         I 


when  the  weight  is  at  B,  is  equal  or  proportional  to  BC  (Fi^r 

169),  wliich   is   evidently  greater  than  GH,  representing  the 

shearing  force  at  />,  when  the  weight  is  at  any  other  point  G. 

Again,  the  bending  moment  at  B  (Fig.  170),  when  W  is  at 

/—  X 
B,  is  W — -. — X.     If  IV  is  at  any  other  point  G  distant  a  from 

O,   the    bendmg    moment    at    B    is    wa — -. —    or    Wx — j— , 

according  as  rz  <  or  >  x,  and  in  either  case  is  greatest  when 
a  =.  x,  i.e.,  when  the  weight  is  at  B. 

Cor.  2.  In  addition  to  the  rolling  load,  let  the  girder  carry 
a  permanent  weight  W  at  the  centre. 

Consider  one  half  of  the  girder  only,  and,  for  convenience 
trace  the  shearing-force  and  bending-moment  diagrams  for  W 
below  OA. 

The  compound   diagram  for  maximum  shearing  forces  is 

W 
DTLFD  (Fig.  171),  where  KT  is  equal  or  proportional  to  — , 


and  KL  =  OF  is  equal  or  proportional  to 


W 


The  maximum  shearing  force  at  a  point  distant  x  from  the 
centre  is  represented  hy  X Y  —  -—[ — [-  .1^)  -| . 


V 

^""^^^ 

■^T 

0 

1 

!k 

\ 

1             i 

1      1 

!       !. 

F 

Y 

L 

c 

A 
/ 

/      1 

/      1 

D 

A 

1 

Y\ 

Fig.  171. 


Fig.  17a. 


Again,  the  compound  diagram  for  maximum  bending  mo- 
ments is  OEFO  (Fig.  172),  where  DF  is  equal  or  proportional 

W'l 
to  ,  and  OF  is  a  straight  line. 


EFFECT  OF  A   ROLLING  LOAD. 


113 


u;  i 


The  maximum  bending  moment  at  a  point  distant  x  from 
the  centre  is  represented  by 

WIP  _ 
I  V  ~ 


XY  = 


+ 


'f(^4 


Cor.  3.   Theoretically,  the  total  volume  of  material  required 
in  the  web  of  the  girder  in  Cor.  2  is  equal  or  proportional  to 


2  X  ar&ixDTLF 


3  Wl      I  W'l 


/,  being  the  web  unit  stress. 

So,  if  d  be  the  effective  depth  of  the  girder,  and  /  the  unit 
stress  in  one  of  the  flanges,  the  total  volume  of  metal  in  that 
flange  is  equal  or  proportional  to 

2  X  area  OEFO       2  WP   ,    WT       i  Wl'   .    i  W'l^ 


+ 


+  « 


fd  ~  lAfd   ^    %fd~6fd^Sfd' 

Case  II.  Let  a  train  weighing  zv  per  unit  of  length  travel 
over  the  girder  from  right  to  left,  and  let  the  total  length  of 
the  train  be  not  less  than  that  of  d 
the  girder. 

The  reaction   at  A,  Fig.   173, 

when  the  front  of  the  train  is  at 

.    wx'  , 

B  distant  x  from  O,  is  — j  ,  and 

is  the  shearing  force  for  all  points 

between  A  and  B.     Upon  the  ver-  F'o-  '73. 

ticals  through  A  and  O  take  AD  and  OE  each  equal  or  pro- 

wl 
portional  to  — .     Thus  between  A  and  B  the  shearing  force 

at  any  point  is  represented  by  the  vertical  distance  between 
that  point  and  a  parabola  having  its  axis  vertical  and  its  vertex 
at  0. 

After  the  end  of  the  train  has  passed  O,  the  shearing  force 
at  any  point  of  the  uncovered  portion  of  the  girder  is  evidently 
represented  by  the  vertical  distance  between  that  point  and  the 
parabola  AFE,  having  its  axis  vertical  and  its  vertex  at  A. 

Again,  as  the  train  moves  from  O  towards  B,  the  reaction 


lii 


;  t  ■  I 


t'l  fi 


'  ii 


1  ! 


!  -; 


114 


THEORY  OF  STRUCTURES. 


\\ 


\\  i 


\     :\  I 


at  Af  and  consequently  the  bending  moment  at  B^  continually 
increase.  On  passing  B^  the  reaction  at  A  still  increases,  and 
the  bending  moment  at  B  when  the  train  covers  a  length  a 
of  the  girder  is 

-^{l-x)  -  -{a  -  xf  =  -^a{2l-a)  -  -— . 

This  expression  is  evidently  a  maximum  when  a{^l  —  a)  is  a 
maximum,  i.e.,  when  a  —  I.  Hence  the  bending  moment,  and 
therefore  the  flange  stresses,  at  any  point  are  greatest  when  the 
moving  load  covers  the  whole  girder. 

Cor.  I.  The  shearing  force  at  any  point  5  is  a  maximum 
when  the  train  covers  the  longest  segment  OB. 

This  is  evidently  the  case  until  the  train  arrives  at  B,  for 
the  reaction  at  A^  and  therefore  the  shearing  force  at  B,  will 
continually  increase  up  to  this  point.  When  the  train  passes 
B  and  covers  a  length  a(^x)  of  the  girder,  the  shearing  force 

at  B  IS  — -. zv{a  —  x). 

1VX 

But  this  is  <  — T- ,  the  shearing  force  at  B  when  OB  is 

2/  ' 


d  —  x^ 
covered,  if   — —, —  <ia  —  x,  i.e, 


..  a-\-  X  

if  — -7— <  I,  which   IS  evi- 


2/     ^  '       ■'  2/ 

dently  the  case. 

Cor.  2.  In  designing  the  flanges  of  a  girder,  the  rolling 
load  is  supposed  to  cover  the  whole  girder,  and  may  be  treated 
as  a  uniformly  distributed  load. 

Cor.  3.  In  addition  tu  the  roll- 
ing load,  let  the  giider  cai^y  a  uni- 
formly distributed  load  of  w'  per 
unit  of  length. 

As  before,  consider  one  half  of 
the  girder  only.     Trace  the  shear- 
ing-force diagram  for  the  perma- 
nent load  below  OA.    The  com- 
pound diagram  is  DHGKy  where  GH  and  AK  are  equal  or 

wl         w'l 
proportional  to  -g-  and  — ,  respectively. 


3 

\vX 

M 

A 

r^>-~->. 

0 

i^?m 

> 

^G 

$ 

y^ 

^ 

y^ 

-. 

Fig.  174. 


■"I:' 


EFFECT  OF  A   ROLLING  LOAD. 


nS 


The  maximum  shearing  force  at  a  point  distant  x  from  the 
centre  is  represented  by  XY  and  is  equal  to 


f;(^+')"+ 


W  X. 


Again,  the  maximum  flange-stresses  are  obtained  by  assum- 
ing the  total  load  upon  the  girder  to  be  zf  4-  w'  per  unit  of 
length.  -     ... 

Ex.  The  two  main  girders  of  a  single-track  bridge  are  80 
ft.  in  the  clear  and  10  ft.  deep.  The  dead  load  upon  the 
bridge  is  2500  lbs,  per  lineal  foot.  If  the  bridge  is  traversed  by 
a  uniformly  distributed  live  load  of  3000  lbs.  per  lineal  foot, 
determine  the  maximum  bending  moment  and  shearing  force 
at  a  point  of  the  girder  distant  10  ft.  from  one  end. 

The  bending  moment  at  any  point  is  a  maximum  when  the 
train  covers  the  whole  of  the  bridge,  in  which  case  the  total 
distributed  load  is  5500  lbs.  per  lineal  foot,  of  which  each  girder 
carries  one  half. 

Thus  the  reaction  at  each  support  =  — .  80. =  i  io,ccX) 

lbs.,  and  t\\Q  bending  moment  at  the  given  point  =  iiooooX  10 
—  10  X  2750  X  5  =  962,500  Ib.-ft. 

The  shearing  force  at  the  given  point  due  to  the  dead  load 
=^  I  loooo  —  10x2750  =  82,500  lbs. 

Tlie  shearing  force  due  to  the  live  load  is  a  maximum,  when 
the  live  load  covers  the  70  ft.  segment,  and  its  value  is  then 


1500x70' 
2x80 


=  45.937ilbs. 


Hence  the  total  maximum  shearing  force 


=  82,500  +  45, 937i  =  128,437^  lbs. 


u\\ 


,  f ,,  i^ 


;%■ 


■Ail 


1 


■"■HRin? 


t 


i:  ' 


1 1 


ii6 


THEORY  OF  STRUCTURES. 


5.  Moments  of  Forces  with  respect  to  a  given  Point  Q. 

— First,  consider  a  single  force  P,. 

Describe  the  force  and  fu- 
nicular polygons,  i.e.,  the  line 
5,5„  and  the  lines  AB,  BC. 

Through  the  point  Q  draw  a 
line  parallel  to  S^S^,  cutting  the 
lines  AB  and  CB  produced  in  x 
and  J. 

Drop  the  perpendiculars  EM 
and  ON  upon  yx  and  5,5,  produced.     Then 

BM  ~  0N~  ON' 
.:  P,BM  =xy.ON. 

But  BM  is  equal  to  the  length  of  the  perpendicular  from  Q 
to  the  line  of  action  of  /*, ,  and  the  product  xy .  ON  is,  there- 
fore, equal  to  the  moment  of  P^  with  respect  to  Q.  Hence,  if 
a  scale  is  so  chosen  that  ON  =.  unity,  this  moment  becomes 
equal  to  xy ;  i.e.,  it  is  the  intercept  cut  off  by  the  two  sides  of 
the  funicular  polygon  on  a  line 
drawn  through  the  given  point 
parallel  to  the  givcfi  force. 

Next,  let  there  be  two  forces, 
P    P  . 

Describe  the  force  and  fu- 
nicular polygons  5j5,5,  and 
ABCD. 

Let  the  first  and  last  sides 
{AB  and  DC)  be  produced  to 
meet  in  G,  and  let  a  line  through 
the  given  point  Q  parallel  to  the  line  5,5,  intersect  these  lines 
in  X  and  y. 

Draw  GM  perpendicular  to  xy,  and  ON  perpendicular  to 
5,5^.    Then 

xy        SjS,  _  resultant  of  /*,  and  P, 
Gll^  ON  ^  "ON  ' 


Fig.  176. 


II' 


r'F" 


MOMENTS  OF  FORCES. 


117 


and  henc6 


(the  resultant  of  P,  and  P^  X  GM  —  xy .  ON. 

But  GM  is  equal  to  the  length  of  the  perpendicular  from  Q 
ipon  the  resultant  of  /*,  and  P^,  which  is  parallel  to  S^S,  and 
must  necessarily  pass  through  G.  Hence,  if  a  scale  is  so  chosen 
that  ON  =^  unity,  xy  is  equal  to  the  moment  of  the  forces  with 
respect  to  Q  ;  i.e.,  it  is  the  intercept  cut  off  by  the  first  and  last 
sides  of  the  funicular  polygon  on  a  line  drawn  through  the  given 
point  parallel  to  the  resultaftt  force. 

A  third  force  P,  may  be  compounded  with  /*,  and  /*, ,  and 
the  proof  may  be  extended  to  three,  four,  or  any  number  of 
forces. 

The  result  is  precisely  the  same  if  the  forces  are  parallel. 

The  force  polygon  of  the  n  parallel  forces  P^,  P^,  .  .  .  P„ 


Fio.  177. 

becomes  the  straight  line  S^S^S^ .  .  .  5„ .  Let  the  first  and  last 
sides  of  the  funicular  polygon  meet  in  G.  Drop  the  perpen- 
iliculars  GM,  ON  upon  xy  and  S,S„,  xy,  as  before,  being  the 
intercept  cut  off  on  a  line  through  the  given  point  Q  parallel 
to  S^S„ .     Then 

xy .  ON  =  GM.  S^S„ .     Hence,  etc. 
Thus  the  moment  of  any  number  of  forces  in  one  and  the 
same  plane  with  respect  to  a  given  point  may  be  represented 
by  the  intercept  cut  off  by  the  first  and  last  sides  of  the  funicu- 


i     1  >' 


f  V     ) 


\'i 


i  i. 


t 


-s 


'i  I  r 


.!'■  f 


H:    ■ 


ii8 


THEORY  OF  STRUCTURES. 


lar  polygon  on  a  line  drawn  through  the  given  point  parallel  to 
the  resultant  of  the  given  forces. 

6.  Bending  Moments. — Stationary  Loads. — Let  a  hori- 
zontal beam  AB,  supported  at  A  and  B,  carry  a  number  of 
weights  -P,,  /*,,  /',,...  at  the  points  iV,,  N,,  N,, ...  - 


Fronr 


Fig.  178. 

The  force  polygon  is  a  vertical  line  1234  .  .  .  n,  where 
12  =  /',,  23  =  P„etc. 

Take  any  pole  O  and  describe  the  funicular  polygon 
A,A^A^  .  .  . 

Let  the ^rst  and  /ast  sides  of  this  polygon  be  produced  to 
meet  in  G  and  to  cut  the  verticals  through  A  and  B  in  the 
points  Cand  D.  - 

Join  CD, 

Let  the  vertical  through  G  cut  AB  in  L  and  CD  in  AT;  LG 
is  the  line  of  action  of  the  resultant. 

Draw  OH  parallel  to  CD. 

From  the  similar  triangles  OvH  and  GCK, 

■  '        0H~  CK' 


R. ,  K  be 
But  L 

Henci 

Thus 
//«r  CD  d 
one  is  equi 
Let  it 
point  Mo 
side  of  M 
In  the 
correspont 
sides  of  th 
OH,  and  y 
these  sides 
the  forces 
bending  m 
from  O  up 
Hence, 
unity,  the  b 
ccpt  071  the 
CD  and  tht 
7-  Mov 
action  of  m 
of  a  railwa) 
determine  t 
loads.     It  I 
wheels  whi( 
apart. 

At  any  \ 


BENDING  MOMENTS. 

From  the  similar  triangles  OnH  and  GDK, 

nH       GK 


119 


OH  ~  DK 


\H     DK 
nH 


BL      R, 


CK  ~  AL 


r: 


R, ,  R,  being  the  reactions  at  A  and  B,  respectively. 
But  iH-\-  nH  =  m  ='P,  -\- P^-\- .  .  .  =  R^-^  R^. 


Hence 


iH=R,     and 


nH=R^. 


Thus  the  line  drawn  throtigh  the  pole  parallel  to  the  closing 
line  CD  divides  the  line  of  loads  into  two  segments,  of  which  the 
one  is  equal  to  the  reaction  at  A  and  the  other  to  that  at  B. 

Let  it  now  be  required  to  find  the  bending  moment  at  any 
point  M  of  the  beam,  i.e.,  the  moment  of  all  the  forces  on  one 
side  of  M  with  respect  to  M. 

In  the  figure  these  forces  are R^,  P^,  P^,  P^,  P^,  P^,  and  the 
corresponding  force  polygon  is  //i 23456.  The  first  and  last 
sides  of  the  funicular  polygon  of  the  forces  are  CD  parallel  to 
OH,  and  A^A^  parallel  to  06.  If  the  vertical  through  J/ meet 
these  sides  in  x  and  y,  then,  as  shown  in  Art.  5,  the  moment  of 
the  forces  R^,  /*, ,  P^,  P^,  P,,  P^  with  respect  to  M,  i.e.,  the 
bending  moment  at  M,  =  ON.xy,  ON  being  the  perpendicular 
from  O  upon  iH  produced. 

Hence,  if  a  scale  is  chosen  so  that  the  polar  distance  ON  is 
?aiity,  the  bending  moment  at  any  point  of  the  beam  is  the  inter- 
cept on  the  vertical  through  that  point  cut  off  by  the  closing  line 
CD  and  the  opposite  bounding  line  of  the  funicular  polygon. 

7.  Moving  Loads. — Beams  are  often  subjected  to  the 
action  of  moving  loads,  as,  e.g.,  in  the  case  of  the  main  girders 
of  a  railway  bridge,  and  it  becomes  a  matter  of  importance  to 
determine  the  bending  moments  for  different  positions  of  the 
loads.  It  may  be  assumed  that  the  loads  are  concentrated  on 
wheels  which  travel  across  the  bridge  at  invariable  distances 
apart. 

At  any  given  moment,  let  the  figure  represent  a  beam  1 1 


*       ! 


1         "h  'I 


.1  M 


^    rf     ' 


kHA 


^  i  '!l 


A  .1! 


ill 


120 


THEORY  OF  STRUCTURES. 


under  the  loads  Z', ,  /*, ,  -P, .  .  .  Describe  the  corresponding 
funicular  polygon  CC'C"  .  .  .  D,  the  closing  line  being  CD. 

Let  the  loads  now  travel  from  right  to  left.  The  result  will 
be  precisely  the  same  if  the  loads  remain  stationary  and  if  the 
supports  1 1  are  made  to  travel  from  left  to  right. 

Thus,  if  the  loads  successively  move  through  the  distances 


^^r-?^ 

.-'>''!./ 


Fig.  179. 

12,  23,  34,  . .  .  to  the  left,  the  result  will  be  the  same  if  the 
loads  are  kept  stationary  and  if  the  supports  are  successively 
moved  to  the  right  into  the  positions  22,  33,  44,  .  .  .  Tli?  new 
funicular  polygons  are  evidently  C'C"  .  .  .  D' ,  C"C' '  ,  .  .  D' , 
C"C""  .  .  .  D'",  ...  the  new  closing  lines  being  CD',  C"D", 
CD'",  .  .  . 

The  bending  moment  at  any  point  M  is  measured  by  xy  for 
the  first  distribution,  x'y'  for  the  second,  x"y"  for  the  third, 
etc.,  the  position  of  M  for  the  successive  distributions  being  de- 
fined by  MM'  -  12,  M'M"  =  23,  M"M"'  =  34,  .  .  . 

Similarly,  if  the  loads  move  from  left  to  right,  the  result 
will  be  the  same  if  the  loads  are  kept  stationary  and  if  the  sup- 
ports are  made  to  move  from  right  to  left. 

It  is  evident  that  the  envelope  for  the  closing  line  CD  for 
all  distributions  of  the  loads  is  a  certain  curve,  called  the  enve- 
lope of  moments.  The  intercept  on  the  vertical  through  any 
point  of  the  beam  cut  off  by  this  curve  and  the  opposite  bound- 


MAXIMUM  SHEAR  AND  BENDING  MOMENT. 


121 


ary  of  the  funicular  polygon  is  the  greatest  possible  bending 
moment  at  that  point  to  which  the  girder  can  be  subjected. 

Example.  Loads  of  12  and  9  tons  are  concentrated  upon  a 
horizontal  beam  of  12  ft.  span  at  distances  of  3  and  9  ft.  from 
the  right-hand  support.    Find  (a)  the  B.  M.  at  the  middle  point 


12' 


9' 


6' 


3' 


I » tons  Tl8ton8-^|    y 

I 


?^ 


Fig.  i8a 

of  the  beam,  and  also  (d)  the  max.  B.  M.  produced  at  the  same 
point  when  the  loads  travel  over  the  beam  at  the  fixed  dis- 
tances of  6  ft.  apart. 

Sca/es  for  lengths,  ^  in.  =  I  ft. ;  for  forces,  ^  in.  =  i  ton. 

Take  polar  distance  =  f  in.  =  ic  tons. 

Case  a.  B.M.  =  ;r>/  X  10  =  3.15  X  10  tons  =  31^  ton-ft. 
Case  b.  B.  M.  —x'y'  X  10  =  3.6    X  10  tons  =  36  ton-ft. 

8.  Analytical  Method  of  Determining  the  Maximum 
Shear  and  Bending  Moment  at  any  Point  of  an  Arbitrarily 
Loaded  Girder  AB. — At  any  given  moment  let  the  load  con- 
sist of  a  number  of  weights  w, ,  w, ,  .  .  .  w„,  concentrated  at 
points  distant  a, ,  rr, ,  .  .  .  a„ ,  respectively,  from  B. 

The  corresponding  reaction  R^  at  a  is  given  by 

R,l  —  %v,a,  +  w^a^  +  .  .  .  +  w^a^, 

/being  the  length  of  the  girder. 

Let  W^  =  zv,-\-'^i-\'  •  '  '  +  ^« >  the  sum  of  the  n  weights. 
"     Wr  —  'w^-\-%v^-\-...-{-Wr^  the  sum  of  the  first  r  w'ts. 
The  shear  at  a  point  P  between  the  rth  and  the  (r  -\-  i)th 
weights  is  ' 

5,  =  /?,  —  w,  —  M/,  —  .  . .  —  w^  =  /?,  —  W'^ . 


<f  ' 


*  t 


U 


'•hir 


I'l  iii 
1  (t 


!    1 


•t  .■>  li 


t    ' 


twwww 


I  u 


il'1 


li^l' 


f  :-:  :. 


122 


THEORY  OF  STRUCTURES. 


Let  all  the  weights  now  move  towards  A  through  a  distance 
X,  and  let  /  of  the  weights  move  off  the  girder,  q  of  the  weights 
be  transferred  from  one  side  of  P  to  the  other,  and  s  new 
weights,  viz.,  w„+, ,  w„+, ,  .  .  .  w„+, ,  advance  upon  the  girder, 
their  distances  from  B  being  a;„+, ,  ^„+, ,  .  .  .  «„+,,  respectively. 

Let  L-=  w^-^-w^-^-  .  .  .  -\-Wp,  the  total  weight  leaving  the 
girder. 

Let  T  =  w4,  +  Wyj^  4"  •  •  •  +  "^r+q ,  the  total  weight  trans, 
ferred  from  one  side  of  P  to  the  other. 

Let  Rpl  =  w,a,  -]-  w^a,-{-  ...-}-  '^p^p . 

"      RJ  =  W„+,rt«+,  +  Zf«+:^„+,  -f  .  .  .   +  W„+.«„+,. 

Thus  Rp,  R^,  R,  are  the  icart''  ns  a'  •'  due,  respectively,  to 
the  weight  which  leaves  the  girder,  the  weight  which  is  trans- 
ferred, and  the  new  weight  which  advances  upon  the  girder. 

The  reaction  R,  at  A  with  the  new  distribution  of  the  loads 
is  given  by 

RJ  =  w^+,Kn  -\-^)  +  •^ph{^p+2  +  ^)  +  .  .  .  +  wXar  +  x) 

.     +  zi>„^,a„+,  =  RJ  -  Rpl  +x{W„-L)  +  RJ, 

and  hence 

(R,  -R,)l=  (/?. -  RpY  +x{W^^  L). 

Also,  the  corresponding  s/iear  at  P  is 

S,  =  R,-  (m/^,  +  m'^+,  -f  .  .  .  +  w,  +  w^,  +  •  •  •  +  Wh^) 
=  7?,-(IK-i^+n      ,    . 

Hence  the  s/iear  at  P  with  the  first  distribution  of  weights 
is  greater  or  less  than  the  s/iear  at  the  same  point  with  the 
second  distribution  according  as 

or  R,~W,>R,-W,-\-L-T, 

or  ..     T-L^R,-  R,, 


M. 


or 


Note.— 
R,,Rp,  an 


according  j 

i.e.,  accord! 
through  w/i 
ivcight  on  t 
Again, 


The  bn 

weights  is 

i]/,  =  RH  - 
=  Ril- 

The  hem 
tribution  is 

M,  =  Rll- 

=  Rli  - 

Hence  tl 
of  weights  ii 
same  point  ^ 

or 

HI-  -2)- 


MAXIMUM   SHEAR  AND  BENDING  MOMENT. 


123, 


or 


T  -  L%R,-  R.^'^i^W^-  L). 


(A). 


Note. — When  no  weights  leave  or  advance  upon  the  girder, 
R,,  Rp,  and  L  are  severally  nil,  and  hence 


according  as 


c  >  c 
•^1  <  'jj »  , 


i.e.,  according  as  the  weight  transferred  divided  by  the  distance 
through  which  it  is  transferred  is  greater  or  less  than  the  total 
zvcight  on  the  girder  divided  by  the  span. 

Again,  let  z  be  the  distance  of  P  from  B,  and  let 

'  • 

R^l  =  zv,a^  -\-  w^a^  -{-...  ■\-  w^a^ . 

The  bending  moment  at  P  with  the  first  distribution  of 
weights  is 

il/,  =  R,{1  —  Z)  —  W^a^  —  2)  —  Ws(^a  —  2)  —  .  .  .   —  W^ttr  —  Z) 

=  Ril-z)-RJ-[-zW,. 

The  bending  moment  at  the  same  point  with  the  second  dis- 
tribution is 

M,  =  R,{1  —s)—  w^,(a^.  -\-x-z)-  w^+3^+,  -\-x  —  z)—... 
—  w^itr  -\-x  —  z)  —  .  .  .  —  %v^^^{arJ,^  -\-x  —  z) 
=  Rll-z)  -  {RJ-Rpl+RJ)  -{x-  z){Wr  -L+T). 

Hence  the  bending  moment  at  P  with  the  first  distribution 
of  weights  is  greater  or  less  than  the  bending  moment  at  the 
same  point  with  the  second  distribution  according  as 


M,>M,, 


or 


Ril-.z)-Rrl-\-zWr%Rll-z)-.{R,-Rp  +  R,)l 


r^;i  :i; 


M 


'  H 


' ,  ii 


I.  1  !!' 


-'if' 


(Ii 


».  .  ' 


I 


' »  .' 


I  i  ,'> 


t-  .i-'l'lfi 


Ii  •. 


! 


124 
or 


THEORY  OF  STRUCTURES. 


zW,-{R,-R^l^{x-z){Wr^L-\-T)\{R,--R:){J'-z) 


or 


z{L  -T+R.-  R,)^l{R,  -  R.)  +  x{Wr  -L+T) 

>^M-^){IV„-L) (B) 


NoU. — If  no  weights  leave  or  advance  upon  the  girder  R,, 
Rp  and  L  are  severally  nil,  and 


M,%M,, 


according  as 


.       -zT^lR,^x{Wr^T^-fl-z)W^.      , 

If  also  the  point  P  coincide  with  the  rth  weight,  and  the 
distance  of  transfer,  x^  =  a,.  -  ^,.^, ,  then 

RJ,  =  WrJ^^a^^ ,     T  =  w,.^., ,    and    2  =  Ur. 
Hence  J/,  ^  J/, ,  according  as 


or 


l-a<'l   ' 

i.e.,  according  as  t/te  sum  of  the  first  r  weights  divided  by  the 
length  of  the  corresponding  segment  is  greater  or  less  than  the 
total  weight  upon  the  girder  divided  by  the  span. 

If  the  weights  are  concentrated  at  the  panel  point3  of  a 
truss,  the  last  relation  may  be  expressed  in  the  form 

first  (r)  weights  ^  total  weight 

r  panels        "^  total  number  of  panels' 

Example.  A  series  of  loads  of  3000,  23,600,  20,100,  21,700, 


MA 

22,900,  1 8,5 
over  a  truss 
Let  Ap 
panel  poin 
compare  th 
lbs.  has  rea( 
weights  hav 

R.= 

9 

Hence  5",  ^ , 


300 


and 


Let  the  v 

Hence  5,  ^  5 
23600  —  o^ 

and 


Hence  the 
weight  of  300 

Again,  let 
moment  at  / 
when  the  wei 
towards  A. 

First.  2  = 


gjiiiii; 


MAXIMUM  SHEAR  AND   BENDING  MOMENT. 


125 


22,900,  18,550,  18,000,  18,000,  and  18,000  lbs.  travel,  in  order, 
over  a  truss  of  240  ft.  span  and  ten  panels. 

Let  Ap^p^ .  .  .  B  be  the  truss,  p^,  P^,  Pt, . .  .  being  the 
panel  points.  Let  the  loads  travel  from  B  towards  A,  and 
compare  the  shear  in  the  panel/,/,  when  the  weight  of  3000 
lbs.  has  reached  />,  with  the  shear  in  the  same  panel  when  the 
weights  have  advanced  another  24  ft. 

i?.  =  i- .  18550  =  1855  lbs.,    Rp  =  o,    ^  =  --, 

W„  =  91300  lbs.,    Z  =  o,     T  =  3000  lbs. 
Hence  5,  ^  Sj,  according  as  (see  A) 


3000  -  o>r,i855  +  —(91300  -  o)^  10985, 


and 


•  •  Oj  ^^  o*  • 

Let  the  weights  again  advance  24  ft. 

I  X  \ 

R,  =  —  .  18000  =  1800  lbs.,    Rp  =  o,     -  —  — , 
10  /       10 

IF„  =  109,300  lbs,,     Z  =  o,     T  =  23,600  lbs. 
Hence  5",  ^  5, ,  according  as  (see  A) 

23600  —  O^  1800  —  o  +  —(109300  —  o),    or    23600^  12730, 


and 


•  •   0|  ^^  Oq  • 


Hence  the  shear  in  the  panel  p^p^  is  a  maximum  when  the 
weight  of  3000  lbs.  is  at  /», . 

Again,  let  the  3000  lbs.  be  at/< ,  and  compare  the  bending 
moment  at  p^  with  the  bending  moment  at  the  same  point 
when  the  weights  have  advanced  first  24  ft.  and  then  48  ft. 
towards^. 

First.  z=.  \20  ft.,  Z  =  o,  T  —  22,900  lbs.,  R,l=  18000X24, 


mmm'WwTr^ 

1    ||||:f  ' 

V  ■ 

lim 

m 

I  a  ■ 


II 


ii 


126 


THEORY  OF  STRUCTURES. 


Rp  =  0,      i?/  =  22900  X  96,       ;jr  =  24  ft.,      PT^  =  68400  lbs., 
W„  =  145,850  lbs.     . 

Hence  M^'^M^,  according  as  (see  B) 

.120(0  —  22900  +  1800  —  o)  -f  22900  X  96  —  18000  X  24 

4-  24(68400  —  o  +  22900)  ^  -—(240  —  120X145850  —  0), 

240 


or 
and 


1425600^  1750200, 


.-.  M^<M,. 


Second,  z  =  120  ft.,  L  =  3000 lbs.,  T=  18550,  ^/=  o,  R/ 
=  3000  X  216,  72/ =  18550X96,  x  =  24it.,  fF^  =  91,300 lbs., 
W„  =  163,850  lbs. 

Hence  M^^M,,  according  as  (see  B) 

120(3000  -  18550  +  O  —  3000 .  —\  +  240(18550 .  -^  —  0) 
+  24(91300—  3000+  1 85  50)  ^-^-(240—  120X163850  —  3000), 


or 
and 


2155200^  1930200, 


Hence  the  bending  moment  at  />,  is  a  maximum  when  the 
weight  of  3000  lbs.  is  at  /, ,  i.e.,  when  all  the  panel  points  are 
loaded. 

9.  Hinged  Girders. — Any  point  of  a  girder  at  which  the 
bending  moment  is  nil  is  termed  a  point  of  contrary  flexure, 
and  on  passing  such  a  point  the  bending  moment  must  neces- 
sarily change  sign. 

Consider  a  horizontal  girder  resting  upon  supports  at  A,  B, 
C,  D,  and  hinged  at  the  points  E  and  F  in  the  side  spans. 

In  order  that  there  may  be  no  distortion  by  the  turning  of  I 
;the  hinges,  the  latter  must  not  be  subject  to  any  bending] 
action ;  i.e.,  they  must  be  points  of  contrary  flexure. 


HINGED   GIRDERS. 


127 


Let  AE  =  a,EB^  3,  BC  -  c,  CF  —  e,  DF  =  d. 

Let  W„  U\,  1V„  W„  W,  be  the  loads  upon  AE,  EB,  BC 
DF,  FC,  respectively,  and  let  ^r, ,  x^,  x^,  x^,  x^  be  the  several 
distances  of  the  corresponding  centres  of  gravity  from  the 
points  E,  B,  C,  F,  C. 


Fig.  181. 


The  two  portions  AE  and  DF  zx&  evidently  in  precisely  the 
same  condition  as  two  independent  girders  of  the  same  lengths, 
carrying  the  same  loiids  and  supported  at  the  ends.  EF  may 
also  be  treated  as  an  independent  girder  supported  at  B  and  C, 
carrying  the  weights  W^,  W^,  W^,  and  loaded  at  the  cantilever 
ends  E  and  F  with  weights  equal  to  the  reactions  at  E  and  F 
for  the  portions  AE,  Z^/^  assumed  to  be  independent  girders. 

Let  R^,  R^,  R^,  R^  be  the  reactions  at  A,  B,  C,  D,  respec- 
tively.   Then 


?nd 


R,a=  lV,x„ 


R,d=W,x,. 


Hence,  since  R^  and  R^  are  always  positive,  there  can  be  no 
upward  pull  either  at  A  or  D,  and  no  anchorage  will  be  needed 
at  these  points. 

Next,  taking  EF  as  an  independent  girder, 

the  load  atE=lV,-R,=  W^.(i  -  ^)  ; 
«       "     F=  W,-R,=  W^i  -  ^. 


■■■m 


.:  ^^.:  pi 


it 


!   Ill  ' 


mm 


Iff 


i   >'. 


128  THEORY  OF  STRUCTURES 

Take  moments  about  C  and  B.   Then 

-{W,-  R,){b  -\-c)-  Wlx,  +  c)  +  R,c  -  W,x,  +  W,x, 

+  ( W,  -  R,)e  =  o. 

and 

-{W,-  R,)b  -  IV^x,  +  WXc  -  X,)  -  R,c 

-f-  ^^^  +  ^)  +  {K-  ^0(^  +  e)  =  o; 

two  equations  giving  /?,  and  i?,,  since  ^,  and  R^  have  been 
already  determined. 

The  pier  moments  P,  at  B  and  /*,  at  C  are 


P,  =  -  ( ^F.  -  /?,)*  -  W^,  = 


W,-{a  -  X,)  -  W^, 


and 


/>,=:_  (J^,  -  R)e  -  U\x,  =  -  W,-^{d  -  X,)  -  W,x, ; 

their  values  depending  solely  upon  the  loads  on  the  spans  contain- 
ing the  hinges. 

The  bending  moment  at  any  point  in  BC  distant  x  from  B 

=  R^x-{W,-  R,){b  4-x)-  Wlx,  -\-x)-M 
=  P^^x{R,-{-R,-  W,-  W,)-M; 

M  being  the  bending  moment  due  to  the  load  upon  the 
length  x. 

The  shearing-force  and  bending-moment  diagrams  for  the 
whole  girder  can  now  be  easily  drawn. 

For  any  given  loads  upon  the  side  spans,  let  AEH  and  DFL 
be  the  bending-moment  curves  for  the  portions  AB,  CD ;  BH 
and  CL  representing  the  pier  moments  at  B  and  C,  respec- 
tively. The  bending  moments  for  the  least  and  greatest  loads 
upon  BC  will  be  represented  by  two  curves  HKL,  IIK'L,  and 
the  distances  TT',  VV  through  which  the  points  of  contrary 
flexure  must  move,  indicate  those  portions  of  the  girder  which 
are  to  be  designed  to  resist  bending  actions  of  opposite  signs. 


HINGED  GIRDERS. 


129 


Again,  let  the  two  hinges  be  in  the  intermediate  span. 

Let  AB  =  a,BE=i  b,  EF  =  c,  FC  =  e,  CD  =  d. 

Let  W,,  W„  W„  JV,,  W,  be  the  loads  upon  AB,  BE,  EF, 
CD,  CF,  rtively,  and  let  ,r, ,  x^,  x,,  x^,  x^  be  the  several 

distances  ^i  the  corresponding  centres  of  gravity  from  the 
points  B,  B,  F,  C,  C. 


Fig.  i8a. 


EF  evi(^  'tly  may  be  treated  as  an  independent  girder  sup- 
ported at  wo  ends  and  carrying  a  load  W^. 

^^  anu  ^^'^  may  be  treated  as  independent  girders  carry- 
ing  the  loads  W^ ,  W^  and  W^,  W^,  respectively,  and  also  loaded 
at  the  cantilever  ends  E  and  F  with  weights  equal  to  the  reac- 
tions at  E  and  F  due  to  the  load  W^  upon  girder  EF,  which  is 
assumed  to  be  independent.     Thus 


the  load  at  ^  =  PF,-  ; 


: " = '■'.(■  -  7)- 


The  pier  moments  P,  at  B  and  P,  at  C  are 


and 


P.=  W,x,+  wii-'^)e; 


their  values  depending  solely  upon  the  loads  on  the  span  contain- 
ing the  hinges. 


130 


THEORY  OF  STRUCTURES. 


V 


Let  R,,  Rj,  Rg,  R,  be  the  reactions  at  A,  B,  C,  D,  respec- 
tively, and  take  moments  about  the  points  B,  A,  D,  C.    Then 

R,a  -  W,x,  +  [w,x,  +  W,x!^  =  o  =  ^.^  -  W,x,  +  P, ; 


-  R,a  +  W,{a  -  X,)  +  Wla  +  a',)  +  VV,x 


rt  +  3 


=  o; 


R,d -w{i-  p){e  +  d)-  IVXx,  +  a')  -  W,{a  -x,)  =  o: 

-  R,d  -wii-  ^).-  -  W,x,  +  W,x,  =  o 

=  -  R,d  +  W,x,  -  P,. 

R^  and  R,  are  always  positive; 

A',  \s/>ositivi'  or  ncgaiivc  according  as  W,^:,  ^  P, ;  and 

R,  "  "  "  "      W,x,->P,. 

Thus  there  will  be  a  downward  pressure  or  an  upward  pull 
at  each  end  according  as  the  moment  of  the  load  upon  the  ad- 
joining span  is  greater  or  less  than  the  corresponding  pier  mo- 
ment. The  ends  must  therefore  be  anchored  down  or  they 
will  rise  off  their  supports.  ' 

The  shearing-force  and  bending-moment  diagrams  for  the 
whole  girder  can  now  be  easily  drawn. 

Let  HEFL  be  the  bending  moment  curve  for  any  given 
load  upon  the  span  BC,  BH  and  CL  being  the  pier  moments 
at  B  and  C,  respectively. 

The  bending-moment  curves  for  the  least  and  greatest 
loads  on  the  side  spans  may  be  represented  by  curves  A  TH, 
ATM  and  DVL,  DV'L,  and  the  distances  TT,  VV  through 
which  the  points  of  contrarj'  flexure  move  indicate  those  por- 
tions of  the  girder  which  are  to  be  designed  to  resist  bending 
actions  of  opposite  signs. 

Reverse  strains  may,  however,  be  entirely  avoided  by 
making  the  length  of  EF  sufficiently  great  as  compared  with 
the  lengths  of  the  side  spans. 

The  preceding  examples  serve  to  illustrate  the  mechanical 
principles  governing  the  stresses  in  cantilever  bridges. 


I.  A  be 

upon  a  suj 

liDrizontal 

segment. 

Sliow  til 
150  lbs.  is  s 
iiig-force  ai 

2.  A  ma 

anci  the  cigl 
if  the  man  i 

3.  Atiml: 
reartion  at  t 

4.  Two  h 
beam  suppor 
of  the  bolls  V 
weight  whicl 
of  maximum 
same  weight 

5.  In  the 
load  which  tl 

Draw  the 

6.  A  unifc 
1,'roiind  and  tl 
at  Oo°  to  the 
bending  men: 
uniformly  disi 
bLiim  should 
2000  lbs.  may 

Draw  curv 


EXAMPLES. 


in 


EXAMPLES. 


1.  A  beam  20  ft.  long  and  weighing  20  lbs.  per  lineal  foot  is  placed 
upon  a  support  dividing  it  into  segments  of  16  and  4  ft.,  and  is  kept 
horizontal  by  a  downward  force  /'  at  the  middle  point  of  the  smaller 
segment.     Find  the  value  of  /'  and  the  reaction  at  the  support. 

Show  that  the  required  force  /'  will  be  doubled  if  a  single  weight  of 
150  lbs.  is  suspended  from  the  end  of  the  longer  segment.  Draw  shear- 
ing-force and  bending  moment  diagrams  in  both  cases. 

Arts.   1200  lbs. ;  1600  lbs. 

2.  A  man  and  eight  boys  carry  a  stick  of  timber,  the  man  at  the  end 
and  the  eight  boys  at  a  common  point.  Find  the  position  of  this  point, 
if  the  man  is  to  carry  twice  as  much  as  each  boy. 

Ans,  Distance  between  supports  =  ^  length  of  beam. 

3.  A  timber  beam  is  supported  at  the  end  and  at  one  other  point;  the 
rcaftion  at  the  latter  is  double  that  at  the  end.     Find  its  position. 

Ans.  Distance  between  supports  =  ^  length  of  beam, 

4.  Two  beams  ABC,  BCD  are  bolted  at  />  and  C  so  as  to  act  as  one 
beam  supported  at/?  and  D\  AB  —  12  ft.,  BC  —  4  ft.,  CD  =  16  ft.  ;  each 
of  liie  bolls  will  bear  a  bending  moment  of  100  lb. -ft.  Find  the  greatest 
weight  which  can  be  concentrated  o  1  the  portion  BC.  Draw  diagrams 
of  nia-ximum  shearing  force  and  bending  moment  when  a  wheel  of  the 
same  weight  rolls  over  the  beam. 

Ans.  \\^^  lbs. 

5.  In  the  preceding  question  find  the  greatest  uniformly  distributed 
load  which  the  beam  will  bear. 

Draw  the  shearing-force  and  bcr>ding-moment  diagrams. 


Ans. 


ilW  lbs. 


6.  A  uniform  beam  20  1/3  ft.  in  length  rests  with  one  end  on  the 
s,'round  and  the  other  against  a  smooth  vertical  wall ;  the  beam  is  inclined 
at  60°  to  the  vertical  and  has  a  joint  in  the  middle  which  can  bear  a 
bending  moment  of  30,000  lb.  ft.  Find  the  greatest  load  which  may  be 
uniformly  distributed  over  the  beam.  Also  find  how  far  the  foot  of  the 
biani  should  be  moved  towards  the  wall  in  order  that  an  additional 
:ooo  lbs.  may  be  concentrated  at  the  joint. 

Draw  curves  of  shearing  force  and  bending  moment  in  each  case. 

Ans.  8000  lbs. ;  distance  =  10  ft. 


\  >i\. 


1  J  fifiiititrt 


(ili. 


;'•■ 


''« 


-\-   -r 


'!> 


i^ 


"■  \ !  I 


mm 


',     i 


"■I'la 


I  m 


'J' 


132 


THEORY  OF  STRUCTURES. 


7.  A  man  of  weight  W  ascends  a  ladder  of  length  /  which  rests 
against  a  smooth  wall  and  the  ground  and  is  inclined  to  the  vertical  at 
an  angle  a.  The  ladder  has  «  rounds.  Find  the  bending  moment  ;ii 
the  nil  round  from  the  foot  when  the  man  is  on  the/th  round  from  the 
foot.     (Neglect  weight  of  ladder.) 


Ans. 


^'pl  ,..   .    ->i   sm  a. 


C«  +  I)' 

8.  A  regular  prism  of  weight  W  and  length  a  is  laid  upon  a  beam  of 
length  2/(>rt).  If  the  prism  is  so  stiff  as  to  bear  at  its  ends  only,  show- 
that  the  bending  action  on  the  beam  is  less  than  if  the  bearing  were  con- 
tinuous from  end  to  end  oi  the  prism. 


Ans. — 1st.  Max.  B.M. 


2d. 


9.  A  railway  girder,  50  ft.  in  the  clear  and  6  ft.  deep,  carries  a  uni- 
formly distributed  load  of  50  tons.  F'ind  the  maximum  shearing  stress 
at  20  ft.  from  one  end,  when  a  train  weighing  ij  tons  per  lineal  foot  crosses 
the  girder. 

Also  find  the  minimum  theoretic  thickness  of  the  web  at  a  supper. 
4  tons  being  the  safe  shearing  inch-stress  of  the  metal. 

Ans.    i6Jtons;  .195  in. 

10.  A  beam  is  supported  at  one  end  and  at  a  second  point  dividing  its 

length  into  the  segments  ;//  and  ;/.     Find  the  two  reactions.     Also  finrl 

the  ratio  of  m  to  n  which  will  make  the  maximum  positive  moment  equal 

to  the  maximum  negative  moment. 

w  —      ,— 

«'),  -_—  (/«  +  nf  ,  m  :  n  ::  I  +  v  3  :  r  2. 


w 
Ans.  — (;«' 
2m 


2>n 


II.  One  of  the  supports  of  a  horizontal  uniformly  loaded  beam  is  at 
the  end.  Find  the  position  of  the  other  support  so  that  the  straining  of 
the  beam  may  be  a  minimum. 

length 


Ans.  Distance  from  end  support  — 


V2 


12.  A  rolled  joist  17  ft.  long  is  supported  at  one  end  and  at  a  point 
13  ft.  distant  from*  that  end.  Two  wagon-wheels  5  ft.  apart  and  eacli 
carrying  a  load  of  1300  lbs.  pass  over  the  joist.  Find  the  maxiniiini 
positive  and  negative  moments  due  to  these  weights,  and  also  the  corre- 
sponding reactions. 

Ans.  Max.  positive  B.  M.  =  5512J  Ib.-ft. ; 

reactions  =  1550  and  1050  lbs. 
Max.  negative  B.  M.  =  5200  lb. -ft.  ; 

reactions  =  1700  lbs.  and  —  400  lbs. 
or  =  2900  lbs.  and  —  300  lbs. 


Denotin 
diagram  for 

13.  A  u 

supports  so 


14.  Two 
su^  ed  u 
disic  it  X  fr 
each  /ft.  in 
X  must  lie. 


15.  A  uni 
supports  at 
given  point 
OP:OQ::C 

Draw  cur 
points  of  the 

16.  A  bea 
mW  and  «^ 
supports.     S 

according  as 

17-  A  whe 
tlie  wheel  in 
and  find  its  v 


18.  Two  \ 
n\V  x.ox\%  m  b 
the  bending  i 

whose  distant 


/ 


< 


a[^  +  |/. 


'9-  Find  tl 

ai  the  two  enc 
one  end  to  w 


EXAMPLES. 


133 


Denoting  the  distance  from  a  support  by  x,  the  max.  positive  B.  M. 
diagram  for  each  half  of  the  13-ft.  span  is  given  by  Mx  =  100(21  —  2x)x. 

13.  A  uniformly  loaded  beam  rests  upon  two  supports.  Place  the 
supports  so  that  the  straining  of  the  beam  may  be  a  minimum. 

Ans,  Distance  cf  each  support  from  centre  =  /( i ]. 

\  V2/ 

14.  Two  bars  AC,  CB  in  the  same  horizontal  line  are  jointed  at  C  and 
suy  ed  upon  two  props,  the  one  at  A,  the  other  at  some  point  in  CB 
disu  it  X  from  C.  The  joint  C  will  safely  bear  n  Ib.-ft.  ;  the  bars  are 
each  /ft.  in  length  and  w  lbs.  in  weight,  Find  the  limits  within  which 
X  must  lie. 

Wl  ±  2« 


Ans.  I 


ywlT  2«' 


15.  A  uniform  load  PQ  moves  along  a  horizontal  beam  resting  upon 
supports  at  its  ends  A  and  B.  Prove  that  the  bending  moment  at  a 
given  point  O  is  a  maximum  when  PQ  occupies  such  a  position  that 
OP:OQ:\OA  :  OB. 

Draw  curves  of  maximum  shearing  force  and  bending  moment  for  all 
points  of  the  beam. 

16.  A  beam  is  supported  at  the  ends  and  loaded  with  two  weights 

rnlV  and  nW  at  points  distant  a,  b,  respectively,  from  the  consecutive 

supports.     Show  that  the   bending  action   is  greatest  at  in  IV  or  nW 

A-  fnyb 

accordmg  as  -  ^  — . 

17.  A  wheel  supporting  10  tons  rolls  over  a  beam  of  20  ft.  span.  Place 
the  wheel  in  such  a  position  as  to  give  the  maximum  bending  moment, 
and  find  its  value. 

Ans.  At  the  centre  ;  50  ton-ft. 

18.  Two  wheels  a  ft.  apart  support,  the  one  vilV  tons,  the  other 
n  ir  tons,  m  being  >  «,  and  roll  over  a  beam  of  /  ft.  span.  Show  that 
the  bending  moment  is  an  absolute  maximum  at  the  centre  or  at  a  point 


whose  distance  from  the  nearest  support  is 


na 


2{in  +  n) 


according  as 


/  ^  (M  I  4-  V  1.  and  find  its  value  in  each 

>    \         ^    m  +  nj 


case. 


m  Wl         ,       VI  +«,,,(,         na     )  '         , 

Ans. ton-ft.  ;  7-  W  \l \    ton-ft. 

4  4/  (         ;«  -I-  «  \ 

19.  Find  the  max.  B.  M.  on  a  horizontal  beam  of  length  /  supported 
ui  the  two  ends  and  carrying  a  load  which  varies  in  intensity  from  w  at 
one  end  to  w  +  px  at  the  other. 


'  I 


ill 


I ;    If 


134 


THEORY  OF  STRUCTURES. 


').■  S. 


20.  Four  wheels  each  carrying  5  tons  travel  over  a  girder  of  24  ft. 
clear  span  at  equal  distances  4  ft.  apart.  Determine,  graphically,  the 
ma.x  B.  M.  at  8  ft.  from  a  support,  and  also  the  absolute  max.  B.  M.  on 
the  girder. 

Ans.  ^\^  ton-ft.  ;  80  ton-ft. 

21.  Two  wheels  each  supporting  7  tons  roll  over  a  beam  of  ^\  ft. 
span.  Find  the  maximum  bending  moment  for  the  whole  span,  and  also 
the  curve  of  the  maximum  bending  moment  at  each  point  when  the 
wheels  are  4  ft.  apart. 

Ans.  Abs.  max.  B.  M.  =  Yo'-  ton-ft.  at  wheel  at  2\  ft.  from  one 
'  end.     Denoting  the  distance  from  support  by  x,  the  max. 
B.  M.  curve  for  the  first  3^  ft.  is  given  by 

^/:«  =  ij(ii  —  2x)x, 
and  for  the  remaining  4  ft.  by 

M.  =  \\{n-x)x. 

22.  Two  wheels  supporting,  the  one  11  tons,  the  other  7  tons,  travel 
over  a  beam  of  1 2\  ft.  span.  Find  the  maximum  bending  moment  for  the 
whole  span,  and  also  the  curves  of  the  max.  shearing  force  (both  positive 
and  negative)  and  maximum  bending  moment  at  each  point  when  the 
wheels  are  6  ft.  apart. 

Ans.  Abs.  max.  B.  M.  =  37.2  ton-ft. 
I  The  max.  positive  shearing  force  at  each  point  is  given  by 

the  equations 


S,= 


183—  iS.r 


and    S.t  = 


7(12^ -.r) 

I2i  ' 


The  max.  negative  %\\e&x'\n^  force  at  each  point  is  giv^en  by 
the  equations 


s  --2^ 


5,= 


45i 


iS.r 


S.  =  - 


42  +  i8.r 


S.= 


I2i'         -  12i  '      -  I2i 

The  max.  B.  M.  curve  is  given  by  the  equations 


183—  i8a-  ,      ,,        ii(i2i 

^  x     and    Afx  =  - 


JXX 


-V) 


124 


I2i 


N.B. — In  the  above  cases  x  is  measured  from  the  support  to  the 
nearest  load. 

23.  In  the  preceding  questio.i  show  that  the  maximum  negative  ?,\\ta.r 
at  4t^  ft.  from  a  support,  when  the  7-ton  wheel  only  is  on  the  beam,  is  the 
same  as  the  maximum  negative  shear  at  the  same  point  when  botli  of 


the  wheels 
maximum 
the  ii-ton 
beam,  and 

24.  Sol 
1250  lbs.  (: 

Ans. 


25.  Thre 
roll  over  a  b 
give  the  ma: 

26.  Place 
any  point  b( 
its  value. 
give  the  sam 
value. 


27.  Four 
roll  over  a  be 
give  the  max 

Am 


28.  All  tL 
B-  M.  at  the  c 
that  for  a  par 
the  beam, 
wheels. 


EXAMPLES. 


135 


the  wheels  are  on  the  beam,  and  find  its  value.  Also  show  that  the 
maximum  negative  shear  at  ^  ft.  from  a  support  is  the  same  when  only 
the  ii-ton  wheel  is  on  the  beam  as  when  the  two  wheels  are  on  the 
beam,  and  find  its  value. 

Ans.  If  J  tons  ;  -\V^  tons. 

24.  Solve  question  22  when  the  beam  carries  an  additional  load  ot 

1250  lbs.  (=  %  ton)  per  lineal  foot. 

/      7418^      \ 
Ans.  Abs.  max.  B.  M.  is  at  5.284  ft.  I  = ft.  1  from  support. 

Max.  positive  shearing-force  diagram  is  given  by  Sx  = 
18.54625— 2.065^:  from  -«r==otojir=6ift.,  and  5^=14.90625— 
1.505.^  from  X  —  6\  to  ;r=  12^  ft.  The  max.  negative 
shtaring-force diagram  is  given  by  5«  =:  —  .56^:  from  x  =  0 
to  X  =  \i\  ft.;  =  3.64  —  1.44^  from  x  =  4/^  tox  =  6i  ft.  ; 
=  '/.54625  —  2.065^  from  .r  =  6J  to  jt  =  6^  ft.;  "^  3.90625  — 
1.505X  from  -r  =  6J  to;r  =  9jit.;  =  9.18625  —  2.065.V  from 
jr  =  9f  to  X  =  I2|  ft.  Max.  B.  M.  curve  is  given  by  j]/x  = 
(18.54625  —  1.7525^).^,  and  Mx  =  (14.90625  —  i.i925.r).r. 

25.  Three  wheels,  each  loaded  with  a  weight  IV  and  spaced  5  ft.  apart, 
roll  over  a  beam  of  12  ft.  span.  Place  the  wheels  in  such  a  position  as  to 
give  the  maximum  bending  moment,  and  find  its  value. 

Ans.  Middle  weight  at  centre  of  beam  ;  4  IV. 

26.  Place  (rt)  the  v/heels  in  the  preceding  question  so  that  B.M.  at 
any  point  between  the  two  hindmost  wheels  may  be  constant,  and  find 
its  value.  Also  (d)  determine  all  the  positions  of  the  wheels  which  will 
give  the  same  bending  moment  at  6  and  12  ft.  from  one  end,  and  find  its 
value. 

Am: — (a)  ist  wheel  at  i  ft.  from  support ;  B.  M.  =  7  IV. 
(d)  When   distance    between   end  wheel  and    sup- 
port is  ^  2  ft.  and  ^  5  ft.;  B.  M.  =  7  IV. 

27.  Four  wheels  each  loaded  with  a  weight  IV  and  spaced  5  ft.  apart 
roll  over  a  beam  of  18  ft.  span.  Place  the  wheels  in  such  a  position  as  to 
give  the  maximum  bending  moment,  and  find  its  value. 

Ans.  One  wheel  off  the  beam  and  middle  wheel  of  remaining 
three  at  the  centre  ;  max.  B.  M.  =  8J  IV.  If  all  wheels 
are  on  beam,  max.  B.  M.  =  8  IV. 

28.  All  the  wheels  in  the  preceding  question  being  on  the  beam,  the 
B.  M.  at  the  centre  for  a  certain  range  of  travel  is  constant  and  equal  to 
that  for  a  particular  distribution  of  the  wheels  when  only  three  are  on 
the  beam.      Find  the  range,  the  B.  M.,  and  the  position  of  the  three 

wiieels. 


d 


k_ 


mti 


;--*^*iriaAii 


^l*''^^ 


1^ 


136 


THEORY  OF  STRUCTURES. 


Ans.  While  the  end  wheel  travels  3  ft.  from  the  support ;  8  W; 
first  wheel  5  ft.  from  the  support. 

29.  A  span  of  /  ft.  is  crossed  by  two  cantilevers  fixed  at  the  ends  and 
hinged  at  the  centre.  Draw  diagrams  of  shearing  force  and  bending 
moment  (i)  for  a  single  weight  W  2X  the  hinge,  (2)  for  a  uniformly  dis- 
tributed load  of  intensity  iv. 

Ans.  Taking  hinge  for  origin,  the  shearing-force  and  bending- 
nioment  diagrams  are  given  by 


(I) 


M:c-=- 


An 


(3) 


Sx  =  iDX ;     Mx 


ivx' 


,@  i  •, 


30.  A  beam  for  a  span  of  100  ft.  is  fixed  at  the  ends.  Hinges  are  in- 
troduced at  points  30  ft.  from  each  end.  Draw  curves  of  shearing  force 
and  bending  moment  (i)  when  a  weight  of  5  tons  is  concentrated  on 
each  hinge  ;  (2)  when  a  uniformly  distributed  load  of  \  ton  per  linear  foot 
covers  (a)  the  centre  length,  {b)  the  two  side  lengths,  ic)  the  whole  span. 

Ans.  Take  a  hinge  as  origin  ;  the  diagrams  are  given  by — 
(i)    For  each  side  span       Sx  =■  I,  Mx=- — 'ix; 

for  centre  span      Sx  =  o,  Mx  =  o. 


(2) — (a)  For  each  side  span  Sx  = 


5 


5 


5 


Af.r=--x: 


for  centre  span     Sx  =  — 


x 
8' 


Mx=^^x- 


i6" 


15        x"" 


ip)  For  each  side  span  5jr  =  „-,  Mx  - 

for  centre  span     Sx  =  o,  Mx  =  o. 

t.       X                       2^         x^ 
(c)  For  each  side  span  Sx  =  —  -t-  V .  ^x  = x  +  -> ; 


for  centre  soan 


■^■'-2       8 


Mx  =  —x 2  • 

2         i6 


31.  If  the  load  on  each  of  the  wheels  in  question  27  is  5  tons,  and  if 
the  beam  also  carries  a  uniformly  distributed  load  of  20  tons,  and  two 
loads  of  2  and  3  tons  concentrated  at  points  distant  5  and  9  ft.,  respec- 
tively, from  one  end,  find  the  maximum  shearing  force  (both  positive 
and  negative)  and  the  maximum  bendmg  moment  for  the  whole  span; 
also  tind  the  loci  for  the  maximum  shearing  force  and  bending  moment 
at  each  point. 


32.  A  rol 

carries  a  uni 
apart,  the  on 
joist.  Find  t 
ft.  from  each 

33-  A  bea 
of  rn  IV  lbs.  a 
from  the  righ 

the  weights  i 

How  will  the 

34-  A  roll< 

carries  the  fou 

end.     Find  th 

both  positive  j 

Ans. 

35-  Solve  I 
lineal  foot  is 
wheels. 

36.  The  loa 

a  beam   of  6c 
16,900,  16,900, 


EXAMPLES. 


137 


Ans.  Denoting  the  distance  from  support  by  x,  the  max. 
positive  shearing-force  diagram  is  given  by  Sx  = 
^  —  ^x  from  ^  =  o  to  JT  =  3 ;  5i-  =  V"  —  \x 
from   jr  =  3  to   .r  =  8  ;  5j:  =  -i^/  —  \x   from   ;ir  =  8   to 


5-^ 


to  x=\Z  ft.     The 


^  =  13  ;  "^-^  =  5  —  78  ^'■o'"  •*'  =  '3 

max.   negative    shearing-force   diagram    is    given    by 

Sx  =  —  -^^x  from  X  -=0  to  x  =  5 ;  5j:  =  f f  —  |jr  from 

;r  =  5  to  .1'  =  10 ;  ^.i-  =  'i.'  —  1-^  from  -r  =  10  to  x  =  1 5  ; 

50       20.r 
5j:  =  -7- -g-  from  .r  =  15  to;r  =  18,     Max.  positive 

/  shear  =  -'/  tons ;  max.  negative  shear  =  ^  tons ;  max. 
bending-moment  curve  is  given  by  Mx  =  --^-x  —  ^§-x'^ 
from  X  =  o  to  .r  =  3  ;  ^/.^  =  ^^x  —  4f.r*  from  .r  =  3  to 
;>r  =  5  ;  Mx^  H^x  —  ^'.r'  —  1 5  from  .r  =  5  to  jr  =  8  ; 
.^/!r  =  ^^i-x  —  \\x''-\-  12  from  jr  =  8  to  JT  =  9 ;  abs.  max. 
B.  M.  =  142  ton. -ft. 

32.  A  rolled  joist  weighing  150  lbs.  per  lineal  foot  and  20  ft.  long 
carries  a  uniformly  distributed  load  of  6000  lbs.,  and  two  wheels  5  ft, 
apart,  the  one  bearing  5000  lbs.  and  the  other  3000  lbs.,  roll  over  the 
joist.  Find  the  maximum  shears  at  the  supports,  at  the  centre,  and  at  5 
ft.  from  each  end. 

Ans.  10,250  lbs. ;  9750  lbs. ;  3250  lbs. ;  6750  lbs.;  6250  lbs. 

33.  A  beam  lit.  long  and  weighing  w  lbs.  per  lineal  foot  has  a  load 
of  m  W  lbs.  at  a  ft.  from  the  left  end  and  a  load  oi  nW  lbs.  at  b  ft. 
from  the  right  end.     Find  the  shearing  forces  and  bending  moments  at 

the  weights  and  at  the  middle  of  the  beam,  a  and  b  being  each  <  — . 

/ 
How  will  the  result  be  affected  '\{b>  —  ? 

'J. 

34.  A  rolled  joist  weighing  450  lbs.  pe*"  lineal  foot  and  20  ft.  long 
carries  the  four  wheels  of  a  locomotive  at  3,  8,  13,  and  18  ft.  from  one 
end.  Find  the  maximum  bending  moment  and  the  maximum  shears, 
both  positive  and  negative,  the  load  on  each  wheel  being  10,000  lbs. 

Ans.  Max.    B.  M.  =  102,000  lb. -ft. ;  max.    shears  =  19,000  lbs. 
and  21,000  lbs. 

35.  Solve  the  preceding  question  when  a  live  load  of  2\  tons  per 
lineal  foot  is  substituted  for  the  four  concentrated  weights  on  the 
wheels. 

36.  The  loads  on  the  wheels  of  a  locomotive  and  tender  passing  over 
a  beam  of  60  ft.  span  are  14,180,  14,180,  21,260,  21,260,  21,260,  21,260, 
16,900,  16,900,  16,900,  16,900  lbs.,  counting  in  order  from  the  front,  the 


M 


,    )         '^ 


4^  -u 


f 


r 

.1  '  ^    Hi 
5       '  J 


[ 


i 


I 

f  V 


¥ 


■  i^  i 


i-  ■" 


if 

III': : 


i;  ;,  . 


t 

_ _    , 

1 

11 

138 


THEORY  OF  STRUCTURES. 


intervals  being  5,  5^,  5,  5,  5,  8|,  5,  4,  5  ft.     Place  the  wheels  in  such  a 
position  as  to  give  the  maximum  bending  moment,  and  find  its  value. 

Also  find  the  maximum  bending  moments  for  spans  of  30,  20,  and  16 
feet. 

Ans.  For  60-ft.  span,  max.  B.  M.  is  at  5th  wheel  and  =  1,559,925.4 

Ib.-ft.  when  ist  wheel  is  7.95   ft.   from 
support. 
For  30-ft.  span,  max.  B,  M.  at  5th  wheel  when  2d  wheel  is 
.596  ft.  from    support  and  =  436,761.4 
Ib.-ft. 
For  2o-ft.  span,  max.  B.  M.  at  centre  when  3d  wheel  is  2^ 
ft.  from  support  and  =  212,600  Ib.-ft.  = 
max.  B.M.  at  same  point  when  4th  wheel 
is  5  ft.  from  support. 
For_i6-ft.  span,   max.  B.  M.  is  at  5th  wheel  and  =  132,875 
Ib.-ft.  when  4th  wheel  is  5  ft.  from  sup- 
port. 

37.  If  the  60-ft.  beam  in  the  preceding  question  also  carries  a  uni- 
formly distributed  load  of  60,000  lbs.,  find  the  curves  of  maximum 
shearing  force  and  bending  moment  at  each  point. 

38.  If  a  beam  is  supported  at  the  ends  and  arbitrarily  loaded,  show 
that  the  ordinate  at  the  point  of  maximum  moment  divides  the  area  of 
the  curve  of  loads  into  two  parts  which  are  equal  to  the  supporting 
forces.  If  a  and  b  are  the  distances  of  the  centres  of  gravity  of  the  parts 
from  the  ends  of  the  beam,  and  if  W  is  the  total  weight  on  the  beam, 

I 


show  that  the  maximum  bending  moment  is 


W  ■ 


& 


39.  A  span  of  /  ft.  is  crossed  by  a  beam  in  two  half-lengths,  sup- 
ported at  the  centre  by  a  pier  whose  width  may  be  neglected.  The  suc- 
cessive weightson  the  wheels  of  a  locomotive  and  tender  passing  overthe 
beam  are  14.000,  22,000,  22,000,  22,000,  22,000,  14,000,  14,000,  14,000,  14,000 
lbs.,  the  intervals  being  T\,  \\,  4J,  \\,  loj,  5,  5,  5  ft.  Place  the  wheels  in 
such  a  position  as  to  throw  the  greatest  possible  weight  upon  the  centre 
pier,  and  find  the  magnitude  of  this  weight  for  spans  of  (1)  50  ft.;  (2)  25 
ft.;  (3)  20  ft.;  (4)  18  ft. 

40.  Loads  of  3f,  6,  6,  6,  and  6  tons  follow  each  other  in  order  over  a 
ten-panel  truss  at  distances  of  8,  5J,  4*,  and  \\  ft.  apart.  Appiy  the 
results  of  Art.  8  to  determine  the  position  of  the  loads  which  will  give 
the  maximum  diagonal  and  flange  stresses  in  the  third  and  fourth  panels. 

41.  A  truss  of  240  ft.  span  and  ten  panels,  has  loads  of  12^,  10,  12,  11, 
9,  9,  9,  9,  and  9  tons  concentrated  at  the  panel  points.  Find,  by  scale 
measurement,  the  bending  moments  at  the  four  panel  points  which  are 
the  most  heavily  loaded,  and  determine  by  Art.  8  whether  these  are  the 


greatest  be 
travel  over 

42.  Loa( 

beam   of   2 
respectively 
moment  at 
the  given  di 

43-  A  be 
the  intermei 
load  upon  tl 
10  tons  unif( 
FC,  30  tons 
at  the  midd 
FC  =  10  h.; 
points  of  infl 

44-  Four 
girder  of  24  1 
left  support, 
centre  of  th( 
distances  apj 
jected. 

45-  Three 
are  placed  up 
'lie  left  abut 
left  abutment 
weights  trave 
which  the  bet 
Ans. 


EXAMPLES. 


13^ 


greatest  bending  moments  to  which  the  truss  is  subjected  as  the  weights 
travel  over  the  truss  at  the  panel  distances  apart. 

42.  Loads  of  7i,  12, 12, 12,  12  tons  are  concentrated  upon  a  horizontal 
beam  of  25  ft.  span  at  distances  of  18,  108,  164,  216,  and  272  in., 
respectively,  from  the  left  support.  Find,  graphically,  the  bcndiny; 
moment  at  the  centre  of  the  span.  If  the  loads  travel  over  the  truss  at 
the  given  distances  apart,  find  the  maximum  B.  M.  at  the  same  section. 

43.  A  beam  ABCD  is  supported  at  four  points  A,  B,  C,  and  D,  and 
the  intermediate  span  ^C  is  hinged  at  the  two  points  E  and  F.  The 
load  upon  the  beam  consists  of  15  tons  uniformly  distributed  over  AB, 
10  tons  uniformly  distributed  over  BE,  5  tons  uniformly  distributed  over 
FC,  30  tons  uniformly  distributed  over  CD,  and  a  single  weight  of  5  tons 
at  the  middle  point  of  EF.  AB  =  1$  ft.;  BE  =  ^  ft.;  EF=  is  ft.; 
FC  =  10  it. ;  CZ>  =  25  ft.  Draw  curves  of  B,  M.  and  S.  F.,  and  tind 
points  of  inflexion. 

44.  Four  wheels  loaded  with  4,  4,  8,  and  8  tons  are  placed  upon  a 
girder  of  24  ft.  span  at  distances  of  3  in.,  6^  ft.,  8J  ft.,  and  9  ft.  from  the 
left  support.  Find  by  scale  measurement  the  bending  moment  at  the 
centre  of  the  girder.  If  the  wheels  travel  over  the  girder  at  the  given 
distances  apart,  find  the  maximum  B.  M.  to  which  the  girder  is  sub- 
jected. 

45.  Three  wheels  loaded  with  8,  9,  and  10  tonsand  spaced  5  ft.  apart, 
are  placed  upon  a  beam  of  15  ft.  span,  the  8-ton  wheel  being  3  ft.  from 
the  left  abutment.  Determine  graphically  the  B.  M.  at  6  ft.  from  the 
left  abutment.  Also  find  the  greatest  B.  M.  at  the  same  point  when  the 
weights  travel  over  the  beam,  and  the  ads.  max.  bending  moment  to 
which  the  beam  is  subjected. 

Ans.  47|  ton-ft.;  53I  ton-ft.;  abs.  max,  B.  M.  =  56^^  ton-ft.  at 
2d  wheel  when  ist  is  2^  ft.  from  support. 


'1, 


I  .1 


Mil'    ui 


.  !  ? 


1  > 


■ 

i! 


I 


CHAPTER   III. 
DEFINITIONS  AND  GENERAL  PRINCIPLES. 

I.  Definitions. — The  science  relating  to  the  strength  of 
materials  is  partly  theoretical,  partly  practical.  Its  primary 
object  is  to  investigate  the  forces  developed  within  a  body,  and 
to  determine  the  most  economical  dimensions  and  form,  con- 
sistent with  stability,  of  that  body.  Certain  hypotheses  have 
to  be  made,  but  they  are  of  such  a  nature  as  always  to  be  in 
accord  with  the  results  of  direct  observation. 

The  materials  in  ordinary  use  for  structural  purposes  may 
be  termed,  generally,  so/td  bodies,  i.e.,  bodies  which  offer  an  ap- 
preciable resistance  to  a  change  of  form. 

A  body  acted  upon  by  external  forces  is  said  to  be  strained 
or  deformed,  and  the  straining  or  deformation  induces  stress 
amongst  the  particles  of  the  body. 

The  state  of  strain  is  simple  when  the  stress  acts  in  one 
direction  only,  and  the  strain  itself  is  measured  by  the  ratio  of 
the  deformation  to  the  original  length. 

The  state  of  strain  is  compound  when  two  (or  more^  stresses 
act  simultaneously  in  different  directions. 

A  strained  body  tends  to  assume  its  natural  state  when  the 
straining  forces  are  removed :  this  tendency  is  called  its  elas- 
ticity. A  thorough  knowledge  of  the  laws  of  elasticity,  i.e.,  of 
the  laws  which  connect  the  external  forces  with  the  internal 
stresses,  is  absolutely  necessary  for  the  proper  comprehension 
of  the  strength  of  materials.  This  property  of  elasticity  is  not 
possessed  to  the  same  degree  by  all  bodies.  It  may  be  almost 
absolute,  or  almost  zero,  but  in  the  majority  of  cases  it  has  a 
mean  value.  Hence  it  naturally  follows  that  solid  bodies  may 
be  classified  between  two  extreme,  though  ideal,  states,  viz., 

140 


a  perfectly 

elastic  bod 

forms  exa( 

fectly  soft 

resistance  t 

Bodies  ( 

tioii  under 

2.  Strej 

jected  to  fi\ 

{a)  A  lo 

{/>)  A  lo 

(e)  A  sh 

stress  tendi 

which  it  is  i 

(d)  A  tr 

(e)  A  tw 
Under  a 

elastic  defoi 
formation,  c 
3.  Resrj 
Let  a  straig 
stretched  or 
distributed 
the  line  of  a 
let  /  be  the  < 
formation. 

If  the  tn 
pared  with 
lim'ts,  the  fo 
and  to  the  a: 
these  quantil 


where  ^  is  a 
and  is  called 
dently  the  fc 

elastic  bar  of 


ELEMENTARY  PRINCIPLES  OF  ELASTICITY. 


141 


a  perfectly  elastic  state  and  a  perfectly  soft  state.  Perfectly 
elastic  bodies  which  have  been  strained  resume  their  original 
forms  exactly,  when  the  straining  forces  are  removed.  Per- 
fectly soft  bodies  are  wholly  devoid  of  elasticity  and  ofTer  no 
resistance  to  a  change  of  form. 

Bodies  capable  of  undergoing  an  indefinitely  large  deforma- 
tion under  stress  are  said  to  ht  plastic. 

2.  Stresses  and  Strains. — Every  body  may  be  sub- 
jected to  five  distinct  kinds  of  stresses,  viz. : 

{a)  A  longitudinal  pull,  or  tension. 

{b)  A  longitudinal  thrust,  or  compression'. 

[c)  A  shear,  or  tangential  stress,  which  may  be  defined  as  a 
stress  tending  to  make  one  surface  slide  over  another  with 
which  it  is  in  contact. 

{(i)  A  transverse  stress. 

{c)  A  twist  or  torsion. 

Under  any  one  of  these  stresses  a  body  may  suffer  either  an 
elastic  deformation,  of  a  temporary  character,  or  a  plastic  de- 
formation, of  a  permanent  character. 

3.  Resistance  of  Bars  to  Tension  and  Compression. — 
Let  a  straight  bar  of  homogeneous  material  and  length  L  be 
stretched  or  compressed  longitudinally  by  a  force  P  uniformly 
distributed  over  the  constant  cross-section  A  of  the  bar ;  let 
the  line  of  action  of  P  coincide  with  the  axis  of  the  bar,  and 
let  /  be  the  consequent  extension  or  compression,  i.e.,  the  de- 
formation. . 

If  the  transverse  dimensions  of  the  bar  are  small  as  com- 
pared with  the  length,  experiment  shows  that,  xvitliin  certain 
limits,  the  force  P  is  directly  proportional  to  the  deformation  / 
and  to  the  area  A,  and  inversely  proportional  to  the  length  L, 
these  quantities  being  connected  by  the  relation 

where  ^  is  a  constant  dependent  upon  the  material  of  the  bar 
and  is  called  the  coefficient  or  modnhts  of  elasticity.  It  is  evi- 
dently the  force  which  will  double  the  length  of  a  perfectly 


IrilMP ' "[, 


elastic  bar  of  unit  section. 


Denoting  the  unit  stress  i-jj  by/", 


\    !ii 


!  '■■  *. 


I-        »    jS 


S40  THEORY  OF  STRUCTURES. 

and  the  strain  per  unit  of  length  (-yj  by  A,  the  above  equation 
.  may  be  written 

or  the  unit  stress  =  E  times  the  unit  strain. 

Thus  the  equation  is  the  analytical  expression  of  Hooke's 
law,  that  for  a  body  in  a  state  of  simple  strain  the  strain  is  pro- 
portional to  the  stress. 

The  longitudinal  strain  is  accompanied  by  an  alteration  in 

the  transverse  dimensions,  the  lateral  unit  strain  being  —  — , 

m 

where  ;;/  is  a  coefficient  which  usually  varies  from  3  to  4  for 
solid  bodies  and  is  approximately  4  for  the  metals  of  construc- 
tion. In  the  case  of  india-rubber,  if  the  deformation  is  small, 
in  is  about  2. 

Generally  the  deformation  may  be  calculated  per  unit  of 
original  length  without  sensible  error,  but  for  india-rubber  it  i-; 
more  accurate  to  make  the  calculation  per  unit  of  stretcJud 

length  (=  -7-^)- 

lateral  strain 


\-\-M 


The  ratio  —  = 


m 


longitudinal  strain 


is  called  Poisson's  ratio. 


If  the  transverse  dimensions  of  a  bar  under  compression 
are  small  as  compared  with  the  length  Z,  a  slight  disturbing 
force  will  cause  the  bar  to  bend  sideways,  and  the  bar  will  be 
subject  -d  to  a  bending  action  in  addition  to  the  compression. 
If  the  bar  is  to  be  capable  of  resisting  a  direct  thrust  only, 
the  ratio  of  L  to  its  least  transverse  dimension  should  not 
exceed  a  certain  limit  depending  upon  the  nature  of  the  ma- 
terial. For  example,  experiment  indicates  that  this  limit 
should  be  about  5  for  cast-iron,  10  for  wrought-iron,  7  for 
steel,  and  20  for  dry  timber. 

If  the  temperature  of  the  bar  is  raised  /°,  the  consequent 
strain  is  at,  a  being  the  coefficient  of  linear  dilatation ;  and  a 
stress  EoLf  will  be  developed  if  a  change  of  length  is  pre- 
vented. 


4ii4'«i>ii'fi;j 


SPECIFIC   WEIGHT.— COEFFICIE^TT  OF  ELASTICITY.    143 


4.  Specific  Weight ;  Coefficient  of  Elasticity ;  Limit 
of  Elasticity  ;  Breaking  Stress. — Before  the  strength  of  a 
body  can  be  fully  known,  certain  physical  constants,  whose 
values  depend  upon  the  material,  must  be  determined. 

{(i)  Specific  Weight. — The  specific  weight  is  the  weight  of  a 
unit  of  volume.  The  specific  weights  of  most  of  the  materials 
of  construction  have  been  carefully  found  and  tabulated.  If 
the  specific  weight  of  any  new  material  is  required,  a  conven- 
ient approximate  method  is  to  prepare  from  it  a  number  of 
regular  solids  of  determinate  volume  and  weigh  them  in  an 
ordinary  pair  of  scales.  The  ratio  of  the  total  weight  of  these 
solids  to  their  total  volume  is  the  specific  weight.  It  must  be 
remembered  that  the  weight  may  vary  considerably  with  time, 
etc.;  thus  a  sample  of  greenheart  weighed  69.75  lbs.  per  cubic 
foot  when  first  cut  out  of  the  log,  and  only  57  lbs.  per  cubic 
foot  at  the  end  of  six  months.  When  the  strength  of  a  timber 
is  being  determined,  it  is  important  to  note  the  amount  of 
water  present  in  the  test-piece,  since  this  appears  to  have  a 
great  influence  upon  the  results. 

The  straining  of  a  structure  is  generally  largely  due  to  its 
own  weight. 

The  total  load  upon  a  structure  includes  all  the  external 
forces  applied  to  it,  and  in  practice  is  designated  dead  {perma- 
nent) or  live  {rolling),  according  as  the  forces  are  gradually  ap- 
plied and  steady,  or  suddenly  applied  and  accompanied  with 
vibrations.  For  example,  the  weight  of  a  bridge  is  a  dead 
load,  while  a  train  passing  over  it  is  a  live  load  ;  the  weight 
of  a  roof,  together  with  the  weight  of  any  snow  which  may 
have  acc'inuL.  upon  it,  is  a  dead  load  ;  7t<ind  causes  at  times 
ex  ^'  vibrations  in  the  members  of  a  structure,  and  al- 
.  often  trep  d  as  a  dead  load,  should  in  reality  be  con- 
red  a  live  loau. 

'he  d(  ,id  loads  of  many  structures  (as  masonry  walls,  etc.) 
are  so  grtit  that  extra  or  accidental  loads  may  be  safely  disre- 
garded. In  cold  climates,  gr  t  masses  of  snow  and  the  pene- 
trating effect  of  the  frost  necessitate  very  deep  foundations, 
which  proportionately  increase  the  dead  weight. 

{b)  Coefficient  of  Elasticity. — Generally  speaking,  a  knowl- 
edge of  the  external  forces      ting  upon  a  structure,  discloses 


■-y 


m 


h' 


t?  :i 


t 


' '% 


,      ,  I  if!  K 

N. ,'    'Ml 


^      'If 

ll     !  '»  -It 


S     I! 


M^i' 


;f  ij  :\  iisi  1 


i;  y  •■; 


144 


THEORY  OF  STRUCTURES. 


1    < 


W  \ 


ili  I 


?  n  ita 
\  1  I 


the  manner  of  their  distribution  amongst  its  various  members, 
but  the  deformation  of  these  members  can  only  be  estimated 
by  means  of  the  coefficient  of  elasticity,  which  expresses  the 
relation  between  a  stress  and  the  corresponding  strain. 

In  practice  it  is  usu  Uy  sufficient  to  assume  that  a  material 
is  elastic,  homogeneous,  and  isotropic,  and  its  deformation 
under  stress  may  be  found,  if  the  coefficients  of  elasticity,  of 
form,  and  of  volume  are  known. 

In  a  homogeneous  solid  there  may  be  twenty-one  distinct 
coefficients  of  elasticity,  which  are  usually  classified  under  the 
following  heads : 

(i)  Direct,  expressing  the  relation  between  longitudinal 
strains  and  normal  stresses  in  the  same  direction. 

(2)  Transverse,  expressing  the  relation  between  tangential 
stresses  and  strains  in  the  same  direction. 

(3)  Lateral,  expressing  the  relation  between  longitudinal 
strains  and  normal  stresses  at  right  angles  to  the  strains;  i.e.,  a 
lateral  resistance  to  deformation. 

(4)  Oilique,  expressing  other  relations  of  stress  and  strain. 
If  a  body  is  isotropic,  i.e.,  equally  elastic  in  all  directions, 

the  tiventy-one  coefficients  reduce  to  tivo,  viz.,  the  coefficients 
of  direct  elasticity  and  of  lateral  elasticity.  Such  bodies,  how- 
ever, are  almost  wholly  ideal.  In  a  perfectly  elastic  body  E 
would  be  the  same  both  for  tension  and  'compression.     In  the 

ordinary  ma  rials  of  construction 
it  is  slightly  less  for  compression 
than  for  tension ;  but  if  the  stresses 
do  not  exceed  a  certain  limit  (§  ic), 
page  145),  the  difference  is  so  slight 
that  it  may  be  disregarded. 

The  equation  J  —  EX   may   be 
represented      gwiphically     by     the 
straight  line  MON,  the  ordinate  at 
any    point    representing    the    unit 
P'o-  '83.  stress  required  to  produce  the  unit 

strain  respresented  by  the  correspoi^ding  abscissa. 

The  angle  MOY  =  tan  -' E  -.  tan  -'(-^V 

Coefficients  of  elasticity  mu.st  be  determined  by  experiment. 


The  co( 
and  timber 
material  to 
the  defiecti 
for  blocks  ( 
verse  loadi 
account  of 
dcformatio 
manship  th 

The  tor 
elastic  resis 
vary  from  t 
elasticity. 

{e)  Lim\ 
body  fall  b( 
forces,  will 
sensible  ch 
ment  of  he 
the  forces 
deformatioi 

Such  a  I 
stress  that  ( 
appreciable 

This  is  J 
the  elastic 
degrees  tha 
tion  betwee 
correctly,  as 
mately  prof 
which  the  d 
load.  In  fa 
terial  depen 
subjected  ai 
ample,  in  e> 
tenacity  of 
Wohler  fou 
but  alternat 
repetitions 


LIMIT  OF  ELASTICITY. 


145 


The  coefficients  of  direct  elasticity  for  the  different  metals 
and  timbers  are  sometimes  obtained  by  subjecting  bars  of  the 
material  to  forces  of  extension  or  compression,  or  by  observing 
the  deflections  of  beams  loaded  transversely.  The  coefficients 
for  blocks  of  stone  and  masonry  might  also  be  found  by  trans- 
verse loading ;  they  are  of  little,  if  any,  practical  use,  as,  on 
account  of  the  inherent  stiffness  of  masonry  structures,  their 
deformations,  or  settlings,  are  due  rather  to  defective  work- 
manship than  to  the  natural  play  of  elastic  forces. 

The  torsional  coQ^cxcnt  of  elasticity,  i.e.,  the  coefficient  of 
elastic  resistance  to  torsion,  has  been  shown  by  experiment  to 
vary  from  two  fifths  to  three  eighths  of  the  coefficient  of  direct 
elasticity. 

{e)  Limit  of  Elasticity. — Whe'i  the  forces  which  strain  a 
body  fall  below  a  certain  limit,  the  body,  on  the  removal  of  the 
forces,  will  resume  its  original  form  and  dimensions  without 
sensible  change  (disregarding  any  effects  due  to  the  develop- 
ment of  heat)  and  may  be  treated  as  perfectly  elastic.  But  if 
the  forces  exceed  this  limit,  the  body  will  receive  a  permanent 
deformation,  or,  as  it  is  termed,  a  set. 

Such  a  limit  is  called  a  ///////  of  elasticity,  and  is  the  greatest 
stress  that  can  be  applied  1:0  a  body  without  producing  in  it  an 
appreciable  and  permanent  deformation. 

rhis  is  an  unsatisfactory  definition,  as  a  body  passes  from 
the  elastic  to  the  non-elastic  itate  by  such  imperceptible 
degrees  that  it  is  impossible  to  fix  any  exact  line  of  demarca- 
tion between  the  two  states.  Fairbairn  defines  the  limit  more 
correctly,  as  the  stress  below  which  the  deformation  is  approxi- 
mately proportional  to  the  load  which  produces  it,  and  beyond 
which  the  deformation  increases  much  more  rapidly  than  the 
load.  In  fact,  both  the  elastic  and  ultimate  strengths  of  a  ma- 
terial depend  upon  the  nature  of  the  stresses  to  which  they  are 
subjected  and  upon  X.\\c  frequency  of  their  application.  For  ex- 
ample, in  experimenting  upon  bars  of  iron  having  an  ultimate 
tenacity  of  46,794  lbs.  per  sq.  in.  and  a  ductility  of  20  %, 
Wohler  found  that  with  repeated  stresses  of  equal  intensity, 
but  alternately  tensile  and  compressive,  a  bar  failed  after  56,430 
repetitions  when  the  intensity  was  33,000  lbs.  per  sq.  in. ;  a 


«l 


lll.^iLt 


iii 


146 


THEORY  OF  STRUCTURES. 


second  bar  failed  only  after  19,187,000  repetitions  when  the 
intensity  was  18,700  lbs.  per  sq.  in. ;  while  a  third  bar  remained 
intact  after  more  than  132,000,000  repetitions  when  the  inten- 
sity was  16,690  lbs.  per  sq.  in.  These  experiments  therefore 
indicated  that  the  limit  of  elasticity  ior  the  iron  in  question, 
under  repeated  stresses  of  equal  intensity,  but  alternately 
tensile  and  compressive,  lay  between  16,000  and  17,000  lbs.  per 
sq.  in.,  which  is  much  less  than  the  limit  under  a  steadily  ap- 
plied stress.  Similar  results  have  been  shown  to  follow  when 
the  stresses  fluctuate  from  a  maximum  stress  to  a  minimum 
stress  of  the  same  kind. 

Generally  speaking,  then,  the  limit  of  elasticity  of  a  ma- 
terial subjected  to  repeated  stresses,  is  a  certain  maximum 
stress  below  which  the  condition  of  the  body  remains  unim- 
paired. 

Bauschinger's  experiments  indicate  that  the  application  to 
a  body  of  any  stress,  however  small,  produces  a  plastic  or 
permanent  deformation.  This,  perhaps,  is  sometimes  due  to  a 
want  of  uniformity  in  the  material,  or  to  the  bar  being  not 
quite  straight  initially.  In  any  case,  the  deformations  under 
loads  which  are  less  than  the  elastic  limit,  are  so  slight  as  to  be 
of  no  practical  account  and  may  be  safely  disregarded. 

The  main  object,  then,  of  the  theory  of  the  strength  of 
materials,  is  to  determine  whether  the  stresses  developed  in  any 
particular  member  of  a  structure  exceed  the  limit  of  elasticity. 
As  soon  as  they  do  so,  that  member  is  permanently  deformed, 
its  strength  is  impaired,  it  becomes  predisposed  to  rupture,  and 
the  safety  of  the  whole  structure  is  threatened.  Still,  it  must 
be  borne  in  mind  that  it  is  not  absolutely  true  that  a  material 
is  always  weakened  by  being  subjected  to  forces  superior  to 
this  limit.  In  the  manufacture  of  iron  bars,  for  instance,  eacli 
of  the  processes  through  which  the  metal  passes  changes  its 
elasticity  and  increases  its  strength.  Such  a  material  is  to  be 
treated  as  being  in  a  new  state  and  as  possessing  new  properties. 

The  strength  of  a  material  is  governed  by  its  tenacity  and 
rigidity,  and  the  essential  requirement  of  practice  is  a  tougli 
material  with  a  high  elastic  limit. 

This  is  especially  necessary  for  bridges  and  all  structures 


BREAKING  STRESS. 


147 


liable  to  constantly  repeated  loads,  for  it  is  found  that  these 
repetitions  lower  the  elastic  limit  and  diminish  the  strength. 

In  the  majority  of  cases,  experience  has  fixed  a  practical 
limit  for  the  stresses,  much  below  the  limit  of  elasticity.  This 
insures  greater  safety  and  provides  against  unforeseen  and 
accidental  loads,  which  may  exceed  the  practical  limit,  but 
which  do  no  harm  unless  they  pass  beyond  the  elastic  limit. 

Certain  operations  have  the  effect  of  raising  the  limit  of 
elasticity :  a  wrought-iron  bar  steadily  strained  almost  to  the 
point  of  its  ultimate  strength  and  then  released  from  strain  and 
allowed  to  rest,  experiences  an  elevation  both  of  tenacity  and 
of  the  elastic  limit. 

If  the  bar  is  stretched  until  it  breaks,  the  tensile  strength 
of  the  broken  pieces  is  greater  than  that  of  the  bar.  A  similar 
result  follows  in  the  various  processes  employed  in  the  manu- 
facture of  iron  and  steel  bars  and  wires :  the  wire  has  a  greater 
ultimate  strength  than  the  bar  from  which  it  was  drawn. 

Again,  iron  and  steel  bars,  subjected  to  long-continued  com- 
pression or  extension,  have  their  resistance  increased,  mainly 
because  time  is  allowed  for  the  molecules  of  the  metal  to  as- 
sume such  positions  as  will  enable  them  to  oiTer  the  maximum 
resistance ;  the  increase  is  not  attended  by  any  ap- 
preciable change  of  density. 

Under  an  increasing  stress  a  brittle  material  will 
be  fractured  without  any  great  deformation,  while  a 
tough  material  will  become  plastic  and  undergo  a 
large  deformation. 

(^)  Breaking  Stress. — When  the  load  upon  a 
material  increases  indefinitely,  the  material  may 
merely  suffer  an  increasing  deformation,  but  generally 
a  limit  is  reached  at  which  fracture  suddenly  takes 
place. 

Cast-iron   is   perhaps   the   most   doubtful   of    all 
materials,  and  the  greatest  care  should  be  observed 
in  its  employment.     It   possesses   little  tenacity  or 
elasticity,  is  very  hard  and  brittle,  and  may  fail  sud-    ^'°'  '^*' 
deiily  under  a  shock  or  an  extreme  variation  of  temperature. 
Unequal  cooling  may  predispose  the  metal  to  rupture,  and  its 


w 

ill 

11 

ill 

ll.. 

If  ■?  1 1 
i  ■ 


148 


THEORY  OF  STRUCTURES. 


strength  may  be  still  further  diminished  by  the  presence  of 
air-holes. 

Cast-iron  and  similar  materials  receive  a  sensible  set  even 
under  a  small  load,  and  the  set  increases  with  the  load.  Thus 
at  no  point  will  the  stress-strain  curve  be  absolutely  straight, 
and  the  point  of  fracture  will  be  reached  without  any  great 
change  in  the  slope  of  the  curve  and  without  the  development 
of  much  plasticity. 

Wroiight-iron  and  steel  are  far  more  uniform  in  their  be- 
havior, and  obey  with  tolerable  regularity  certain  theoretical 
laws.  They  are  tenacious,  ductile,  have  great  compressive 
strength,  and  are  most  reliable  for  structural  purposes.  Their 
strength  and  elasticity  may  be  considerably  reduced  by  high 
temperatures  or  severe  cold. 

When  a  bar  of  such  material  is  tested,  the  stress-strain 
curve  (/  =  ±  ETC),  as  has  already  been  pointed  out,  is  almost 
absolutely  straight  within  the  elastic  limit,  e.g.,  from  O  lo  A  in 

tension  and  from  O  to  B  in  com- 
pression. As  the  load  increases 
beyond  the  elastic  limit,  the  in- 
creasing deformation  becomes 
plastic  and  permanent,  and  the 
stress-strain  diagram  takes  an  ap- 
preciable curvature  between  the 
limits  A  and  B  and  the  points  D 
and  E  corresponding  to  the  maxi- 
mum loads.  In  tension,  as  soon 
as  the  point  D  is  reached,  the  bar 
rapidly  elongates  and  is  no  longer 
able  to  sustain  the  maximum  load, 
its  sectional  area  rapidly  dimin- 
ishes, and  fracture  ultimately  takes 
place  under  a  load  much  less  than 
the  maximum  load.  The  point  of 
fracture  is  represented  in  the  figure 
by  the  point  F  the  ordinate   of  F  being  the  actual  Jiltiuiatc 

final  load  on  the  bar 

intensity  of  stress  —  —  ■-  -j—. — j -■ — . 

•^   -^  area  of  fractured  section 


Fig.  iB^. 


vt* 


BREAKING  STRESS. 

The  exact  form  of  the  stress-strain  curve-between  D  and  F 
is  unknown,  as  no  definite  relation  has  been  found  to  exist  be- 
tween the  stress  and  strain  during  the  elongation  from  D  to  F. 

Ordinarily,  the  hrtaking  tensile  stress  has  been  defined  to  be 
the  maximum  lead  applied  divided  by  the  initial  sectional  area 
of  the  bar;  but  this,  although  convenient,  is  manifestly  in- 
correct. 

It  is  important  to  note  that,  as  the  deformation  gradually 
increases  under  the  increasing  load,  the  molecules  of  the  ma- 
terial require  greater  or  less  time  to  adjust  themselves  to  the 
new  condition. 

During  the  tensile  test  of  a  ductile  material  there  is,  at 
some  point   beyond   the  elastic   limit,  an 
abrupt  break  GH  in  the  continuity  of  the 
stress-strain  curve,  the  curve  again  becom- 
ing continuous  from  H  to  D. 

The  point  G  has  been  called  the  Yield 
Point  or  the  Breaking-down  Point,  and  the 
deformation  from  H  onward  is  almost 
wholly  plastic  or  permanent. 

In  compression  there  is  no  local  stretch 
as  in  tension,  and  there  is   consequently  A 
no  considerable  change  in  the  curvature  of  Fig.  ise. 

the  compression  stress-strain  curve  up  to  the  point  of  fracture. 

Timber  is  usually  tested  by  being  subjected  to  the  action 
of  tensile,  compressive,  or  transverse  loads.  Other  character- 
istics, however,  must  be  known  before  a  full  conception  of  the 
strength  of  the  wood  can  be  obtained.  Thus  the  specific 
weight  must  be  found  ;  the  amount  of  water  present,  the  loss 
in  drying,  and  the  corresponding  shrinkage  should  be  deter- 
mined ;  the  structural  d'x^QV&ncQS  of  the  several  specimens,  the 
rate  of  growth,  etc.,  should  be  observed. 

The  chief  object  of  experiments  upon  masonry  and  orick- 
icork  is  to  discover  their  resistance  to  compression,  i.e.,  their 
crushing  strength.  In  fact,  their  stiffness  is  so  great  that  they 
may  be  compressed  up  to  the  point  of  fracture  without  sensible 
change  of  form,  and  it  is  therefore  very  difficult,  if  not  impos- 
sible, to  observe  the  limit  of  elasticity. 


iTTir 


ii 


I  !   * 


.i       I 


150 


THEORY  OF  STRUCTURES. 


The  cement  or  mortar  uniting  the  stones  and  bricks  is  most 
irregular  in  quality.  In  every  important  work  it  should  be  an 
invariable  rule  to  prepare  specimens  for  testing.  The  crushing 
strength  of  cement  and  of  mortar  is  much  greater  than  the  ten- 
sile strength,  the  latter  being  often  exceedingly  small.  Hence  it 
is  advisable  to  avoid  tensile  stresses  within  a  mass  of  masonry, 
as  they  tend  to  open  the  joints  and  separate  the  stones  from 
one  another.  Attempts  are  frequently  made  to  strengthen 
masonry  and  brickwork  walls  by  inserting  in  the  joints  tarred 
and  sanded  strips  of  hoop-iron.  Their  utility  is  doubtful,  for, 
unless  well  protected  from  the  atmosphere,  they  oxidize,  to  the 
detriment  of  the  surrounding  material,  and,  besides  this,  they 
prevent  an  equable  cistribution  of  pressure.  They  are,  how- 
ever, far  preferable  to  bond-timbers. 

The  working  load  (or  stress,  or  strength)  is  the  maximum 
stress  which  a  material  can  safely  bear  in  ordinary  practice, 
and  depends  both  upon  the  character  (see  Art.  5,  below)  of  the 
stress  and  upon  the  ultimate  strength  of  the  material,  the  ratio 
of  the  ultimate  or  breaking  stress  to  the  working  stress  being 
usually  called  a  factor  of  safety.  For  example,  the  factor  is 
about 

3  for  long-span  iron  bridges,  or  bridges  having  great  weight 
as  compared  with  the  live  load  (a  moving  train). 

4  for  ordinary  iron  bridges. 

5  for  ordinary  metal  shafting. 

8,  10,  and  even  more  for  long  struts  and  members  subjected 
to  repeated  stresses  of  varying  magnitude. 

10  is  also  generally  taken  to  be  the  factor  of  safety  for 
timber. 

Under  a  steady,  or  a  merely  statical  load,  even  as  great  as 
|-  of  the  breaking  stress,  a  member  of  a  structure  may  prob- 
ably not  be  unsafe. 

5.  Wohler's  Law. — It  is  now  generally  admitted  that 
variable  forces,  constantly  repeated  loads,  and  continued  vibra- 
tions diminish  the  strength  of  a  material,  whether  they  pro- 
duce stresses  approximating  to  the  elastic  limit,  or  exceedingly 
small  stresses  occurring  with  great  rapidity.  Indeed  many 
engineers  design  structures  in  such  a  manner,  that  the  several 


members  £ 
of  the  evil 
Although  1 
tacitly  ack 
first  to  giv 
vation  and 
"  That 
of  a  bar,  tl 
peated  strt 
differences 
affected  in 
produce  Ixi 
This  •  la 
ments  the 
great  rapid 
resulting  st 
to  the  rapi 
stress,  and 
subject  for 
The  exf 
repetitions  ( 
rapid  than  t 
depends  boi 
cnce  or  flue, 
The  efife 
nately  tensii 
in  Art.  4. 

Bars  of  t 

bore  31,132 

moved  betw 

wlien  the  stt 

The  tabl 

ments  on  ste 

The  axle- 

when  subjecl 

site  in  kind,  ; 

were  of  the 

shearing  stre 


VVOHLER'S  LAW. 


«5i 


members  arc  strained  in  one  way  only,  so  convinced  are  they 
of  tlie  evil  effect  of  alternating  tensile  and  compressive  stresses. 
Although  the  fact  of  a  variable  ultimate  strength  had  thus  been 
tacitly  acknowledged  and  often  allowed  for,  Wohler  was  the 
first  to  give  formal  expression  to  it,  and,  as  a  result  of  obser- 
vation and  experiment,  enunciated  the  following  law : 

"That  if  a  stre^.s  t,  due  to  a  static  load,  cau'  :  the  fracture 
of  a  bar,  the  bar  may  also  be  fractured  by  a  series  of  often-re- 
peated stresses,  each  of  v.'hich  is  less  than  t\  and  that,  as  the 
differences  of  stress  increase,  the  cohesion  of  the  material  is 
affected  in  such  a  manner  that  the  minimum  stress  required  to 
produce  fracture  is  diminished." 

This  -law  is  manifestly  incomplete.  In  Wohler's  experi- 
ments the  applications  of  the  load  followed  each  other  with 
great  rapidity,  yet  a  certain  length  of  time  was  required  for  the 
resulting  stresses  to  attain  their  full  intensity ;  the  influence  due 
to  the  rapidity  of  application,  to  the  rate  of  increase  of  the 
stress,  and  to  the  duration  of  individual  strains  still  remains  a 
subject  for  investigation. 

The  experiments,  however,  show  that  the  rate  of  increase  of 
repetitions  of  stress  required  to  produce  fracture,  is  much  more 
rapid  than  the  rate  of  decrease  of  the  stresses  themselves,  and 
depends  both  upon  the  maximum  stress  and  upon  the  differ- 
ence ox  fluctuation  of  stress. 

The  effect  of  repeated  stresses  of  equal  intensity,  but  alter- 
nately tensile  and  compressive,  has  been  already  pointed  out 
in  Art.  4. 

Bars  of  the  same  material  repeatedly  bent  in  one  direction, 
bore  31,132  lbs.  per  square  inch  when  the  load  was  wholly  re- 
moved between  each  bending,  and  45,734  lbs.  per  square  inch 
when  the  stress  fluctuated  between  45,733  lbs.  and  24,941  lbs. 

The  table  on  page  152  gives  the  results  of  similar  experi- 
ments on  steel. 

The  axle-steel  was  found  to  bear  22,830  lbs.  per  square  inch, 
when  subjected  to  repeated  shears  of  equal  intensity  but  oppo- 
site in  kind,  and  29,440  lbs.  per  square  inch,  when  the  shears 
were  of  the  same  kind.  It  would  therefore  appear  that  the 
shearing  strengths  of  the  metal  in  the  two  cases  are  about  ^ 


152 


THEORY  OF  STRUCTURES. 


of  the  strengths  of  the  same  metal  under  alternate  bending  and 
under  bending  in  one  direction,  respectively. 


Character  of  Fluctuation. 


Alternating  stresses  of  equal  intensity  . . . . 
Complete  relief  from  stress  between   each 

bending 

Partial   relief  from    stress    between   each 

bending 


Maximum  Resistance  to  Repeated 
Stresses  in  lbs.  per  square  inch. 


Axle-steel. 


29,000,-29,000 
49,890,  O 

83,110,      36,380 


.S|prinf;-steel  (un- 
hardened). 


52,000,        0 

93,500,   62,240 


From  torsion  experiments  with  various  qualities  of  steel, 
the  important  result  was  deduced,  that  the  maximum  resistance 
of  the  steel  to  alternate  twisting  was  f  of  the  maximum  resist- 
ance of  the  same  steel  to  alternate  bending.  ' 

Wohler  proposed  2  as  a  factor  of  safety,  and  considered 
that  the  maximum  permissible  working  stresses  should  be  in 
the  ratios  of  1:2:3,  according  as  members  are  subjected  to 
alternate  tensions  and  compressions  (alternate  bending),  to 
tensions  alternating  with  entire  relief,  or  to  a  steady  load. 

The  weakening  of  metal  by  repeated  stresses  has  been 
called /<rz^?^t',  and  is  much  more  injurious  to  iron  and  steel 
under  tension  than  under  compression.  Egleston's  investiga- 
tions have  shown  that  a  fatigued  metal  may  sometimes  be 
restored  by  rest  or  by  annealing. 

From  the  law,  however,  as  it  stands  formulae  may  be  de- 
duced which,  it  is  claimed,  are  more  in  accordance  with  the 
results  of  experiment,  give  smaller  errors,  and  insure  greater 
safety  than  the  false  assumption  of  a  constant  ultimate 
strength. 

The  formulae  necessarily  depend  upon  certain  experimental 
results,  but  in  applying  them  to  any  particular  case,  it  must  be 
remembered  that  only  such  results  should  be  employed,  as 
have  been  obtained  for  material  of  the  same  kind  and  under 
the  same  conditions  as  the  material  under  consideration.  The 
effects  due  to  faulty  material,  rust,  etc.,  are  altogther  indeter- 
minate, so  that  no  formula  can  be  perfectly  universal  in  its 
application.  Hence  the  necessity  for  factors  of  safety,  with 
values  depending  upon  the  class  of  structure,  still  exists. 


A  brief 

now  be  gi\ 

ty  the  SI 

under  a  st£ 

n,  the  p 
a  given  nui 
tion  remaii 
tension,  a  ( 

s,  the  vi 
alternating 
due  to  a  vil 
librium. 

/'  is  the 

F  is  the 


6.  Laun 

subjected  tc 

•  compressive 

mum  rt,  (=  ; 

Let  a,  — 

Let 

If  a^  =  o 
If  d=o 
By  Wohl 

/being  an  ui 
be  determine 

If  ^  =  u, 


« 


LAUNHARDT'S  FORMULA. 


153 


A  brief  description  of  the  principal  of  these  formulae  will 
now  be  given,  and  in  the  discussion 

/,  the  statical  breaking  strength,  is  the  resistance  to  fracture 
under  a  static  or  under  a  very  gradually  applied  load. 

n,  the  primitive  strength,  is  the  resistance  to  fracture  under 
a  given  number  of  repeated  stresses,  the  stress  in  each  repeti- 
tion remaining  unchanged  in  kind,  i.e.,  being  due  either  to  a 
tension,  a  compression,  or  a  shear. 

s,  the  vibration  strength,  is  the  resistance  to  fracture  under 
alternating  stresses  of  equal  intensities,  but  different  in  kind, 
due  to  a  vibratory  motion  about  the  unstrained  state  of  equi- 
librium. 

/'  is  the  admissible  stress  per  unit  of  sectional  area 

F  is  the  effective  sectional  area  and  is 

_  numerically  absolute  maximum  load 
_  _  . 

6.  Launhardt's  Formula. — A  bar  of  unit  sectional  area  is 
subjected  to  stresses  {B)  which  are  either  wholly  tensile,  wholly 
compressive,  or  wholly  shearing,  and  which  vary  from  a  maxi- 
mum rt,  (=  max.  B)  to  a  minimum  «,  (=  min.  B). 

Let  a^  —  a^^=^  d  =.  the  maximum  difference  of  stress. 


Let 

If  a,  =  o, 
If  d=o, 
By  Wohler's  law, 


a^  _  min.  B 
a^       max.  B 

a^  =  a,=  t. 


a,  cxd  =  /d, 


=  0. 


(0 


/being  an  unknown  coefficient  of  which  the  value  remains  to 
be  determined. 


If  rtT  =:  o, 

\i  d  ■=  11, 


a^=-  t    and    /'  =  00. 
a^  ^  d   and    /=  i. 


tel 


—4--  - 


f  Wfr 

I'r 

1 

i  '  "t 

^' 

154  THEORY  OF  STRUCTURES. 

Launhardt's  assumption,  viz.,/= ,  satisfies  these  ex- 

/  —  rt, 

treme  conditions,  and  also  gives  intermediate  values  of  <z,  which 

closely  agree  with  the  results  of  the  most  reliable  experiments. 

Hence  (i)  becomes 

t  —  u         t  —  u ,  . 

a,  —  ' d  — --{a,  —  <?,), 


<?, 


a 


and 


This  is  Launhardt's  formula,  and  is  an  analytical  expression 
of  Wohler's  Law. 

Wohler  in  his  bending  experiments  upon  Phoenix  axle-iioii 
found  that  ti  =  2195*  per  cent.'*  and  t  =  4020*  per  cent.'; 

•  ^~  ^^  =  i 
u  6 

The  same  iron  under  tension  gave  u  =  2195*  per  cent.'  and 
t  =  3290*  per  cent." ; 

.  t  —  u  _  I 
'        u  2 

Choosing  the  most  unfavorable  case,  and,  in  order  to  insure 
greater  safety,  taking  u  =  2100*  per  cent.',  equation  (2)  becomes 


f         0^ 

«,  =  21001^1  +-j. 


(3) 


If  3  is  the  factor  of  safety, 

*  =  7oo(.  +  f). 


(4) 


*  k  per  cent.'^  is  an  abbreviation  for  kilogrammes  per  square  centimetre. 
One  kilogramme  per  square  centimetre  is  equivalent  to  14.2232  lbs.  per  sq.  in. 


LAUNHARDT'S  FORMULA. 


'55 


In  his  bending  experiments  upon  Krupp  cast-steel  (untcm- 
porcd)  it  was  found  that  u  =  3510*  per  cent.'  and  t  =  7340*  per 

cent'. ; 


t  —  ti 


ti 


7 
6' 


But  steel  varies  considerably  in  strength,  and  great  care 
must  be  exercised  in  its  use,  especially  in  bridge  construction. 
For  this  reason  take  u  =  3300*  per  cent."  and  /  =  6000*  per  cent.'; 


/  —  «        Q 


u 


II 


and  (2)  becomes 


a,  =  3300 


i'+U 


.    .    (5> 


If  3  is  the  factor  of  safety, 

3=  1100(1 +-^0) (6) 

Example  i. — The  stresses  upon  a  bar  of  Phoenix  axle-iron, 
normal  to  its  cross-section,  vary  from  a  maximum  tension  of 
50000*  to  a  minimum  tension  of  20000*.  Determine  the  admis- 
sible stress  per  cent."  and  the  necessary  sectional  area. 

By  (4), 

/      ,     I  20OOO\        „      , 
^  =:.  70o(  I  +  -  — -- j  =  840*  per  cent.', 


and 


2  50000 


_       50000        50000 
.*.  F  =  ; —  =     „        =  59.52  sq.  centimetres. 


840 


Let/  be  the  dead  load  and  q  the  total  load,  per  lineal  unit 
of  length,  upon  the  flanges  of  roof  and  bridge  trusses. 


:_ili.-. 


-I   ( 


]■  ^ 


t 


i 


\u 


156  •       THEORY  OF  STRUCTURES. 

P 
.-.  0  =  - ,  and  equations  (4)  and  (6)  respectively  become 

b=    700(1 +i^) (7) 

*=uoo(,+f,|)..     .....    (8) 

Ex.   2. — Determine  the  limiting  stress  per  cent."  for  the 
flanges  of  a  wrought-iron  lattice  girder  when  the  ratio  of  the 

dead  load  to  the  greatest  total  load  is 


By  (7). 


ZV 


^  =  700(1+^-1)  =  800*. 


7.  Weyrauch's  Formula. — Let  a  bar  of  a  unit  sectional 
area  be  subjected  to  stresses  which  are  alternately  different  in 
kind,  and  which  vary  from  an  absolute  numerical  maximum  a 
{=  max.  £)  of  the  one  kind  to  a  maximum  a"  (=  max.  B')  of 
the  other  kind.  f 

Let  a'  -^  a"  =  d  =  the  maximum  numerical  difference  of 
stress. 


=  0'. 


Let 

a         max.  ±( 

a!  ~  max.  B 

If  a"  =  0, 

a'  —  d  —u. 

If  a"  ==  s, 

d 

a  =  s  =  —. 

2 

By  Wohler's 

Law 

'  a'  cxd  =/d, 

(9) 


/  being  an  unknown  coefficient  of  which  the  value  remains  to 
be  determined. 


If  a'  =■  u, 
If  a'  =  s, 


WEYRAUCH'S  FORMULA. 


157 


Weyrauch's  assumption,   viz.,  / 


-,   satisfies 


2u  —  s  —  a' 

these  extreme  conditions,  the  most  reliable  results  of  the  few 
experiments  yet  recorded,  and  also  Wohler's  deduction  that  a' 
diminishes  as  d  increases  and  vice  versa. 
Hence  (9)  becomes 


n  —  s 


a'  = 


2n  —  s  —  a 


id  = 


n  —  s 


211  —  s  —  a 


-{a'  +  a"\ 


and 


/        u  —  sa"\         I        u  —  s   ,\ 
•.a'  =  u[i -j  =  u[i 0'). 


(10) 


This  is  Weyrauch's  formula,  and  it  may  be  always  applied 
to  those  cases  in  which  a  member  is  subjected  to  stresses  alter- 
nating between  tension  and  compression,  or  due  to  shearing 
actions  in  opposite  directions. 

In  the  Phoenix  iron  experiments  already  referred  to  it  was 
found  that  s  =  1 1 70*  per  cent.' ; 


u 


u 


7_ 
15 


Taking  7^  =  2100*  as  before,  and  making 


» 


w 


^_  I 

~  2 


(10) 


becomes 


«'  =  2100(^1 -j.      .....     (11) 

If  3  is  the  factor  of  safety, 

0  =  700I I I 


(12) 


Weyrauch  considers  3  to  be  the  proper  factor  of  safety  for 
bridges  and  similar  structures.  It  is  also  a  suitable  factor  for 
the  parts  of  machines  subjected  to  determinate  straining 
actions.  A  larger  factor  will  be  required  when  other  con- 
tingencies have  to  be  provided  against. 


I  .fi 


^3 


IlM 


158 


THEORY  OF  STRUCTURES. 


i. 


In  the  steel  experiments,  Wohler  found  that  s  —  2050*  per 
cent.' ; 


u  —  s 
u 


_5_ 
12 


Taking  u  =  3300*  and  s  =  1800*, 

u  —  s       5 


u 


II 


and  (10)  becomes 

a' =  3300(1  -  j\(f' } (13) 

If  3  is  the  factor  of  safety, 

d  =  1100(1  -^j<p'\ (14) 

If  a  very  soft  steel  is  employed  in  the  construction  of  a 
bridge,  it  may  be  advisable  to  diminish  still  further  the  ad- 
missible stress  per  unit  of  sectional  area.  For  example,  it  may 
be  assumed  that  /  =  5200*,  u  =  3000*,  and  s  =  1500*,  so  that 
(2)  and  (10)  respectively  become 


and 


a,  =  3000(1  +10) (15) 

a'  =z  3000(1  —  ^cp') (16) 

Example. — The  stresses  in  a  wrought-iron  bar  normal 
to  its  cross-section,  vary  between  a  tension  of  40000*  and  a 
compression  of  30000*.  Find  the  sectional  area  (disregarding 
buckling). 

By  (12) 

.^  =  700(1  -  ^  X  UiU  =  437. 5'  per  cent.\ 

£.      40000  .. 

.'.  F  —  =  91.42  sq.  cei.kimetres. 

437-5       ^   ^     ^ 

Shearing  Stresses. — For  shearing  stresses  in  opposite  direc- 
tions Wohler  fomtd,  in  the  case  of  Krupp  cast-steel  (untem- 


pered),  that 
about  ^  of 
alternately 
assumed,  t: 
value  for  str 
and  which  '. 
8.  Unwi 
stress  in  the 


a  being  the 
statical    bre; 
remains  to  t 
When  d 
When  d 


When  d : 
pressive  and 

a'  to  —  a',  ar 

In  these 
wrought-iron 
identical  witl 
therefore  be 
mediate  case 

The  meai 
that  the  forn 


Example 
<»  sectional  ar 


a 


line 


WEYRAUCH'S  FORMULA. 


m 


pered),  that  ti  =  2780*  per  cent.'  and  s  =  1610*  per  cent.',  or 
about  ^  of  the  corresponding  values  for  stresses  which  are 
alternately  tensile  and  compressive,  and  it  may  be  generally 
assumed,  tl  it  the  value  of  b  for  shearing  stresses,  is  ^  of  its 
value  for  stresses  which  are  alternately  tensile  and  compressive, 
and  which  \iave  the  same  ratio  0'. 

8.  Unwin  has  proposed  to  include  all  cases  of  fluctuating 
stress  in  the  formula 


£1 


a  being  the  actual  strength,  d  the  fluctuation  of  stress,  t  the 
statical  breaking  strength,  and  n  a  coefficient  whose  value 
remains  to  be  determined. 

When  d  =  o,  the  load  is  steady  and  a'  =  t. 

When  d=  a\  the  load  alternates  with  entire  relief  and 


a!  —  2t{\' \ -\- li' —  n). 

When  d  =  2a\  the  stresses  are  alternately  tensile  and  com- 
pressive and  of  equal  intensity.     The   stress  fluctuates   from 

t 
a'  to  —  a' ,  and  «'  =  —  . 

211 

In  these  extreme  cases,  if  n  is  made  equal  to  1.42  for 
wrought-iron  and  to  1.66  for  steel,  results  are  obtained  almost 
identical  with  those  given  in  Arts.  6  and  7.  The  formula  may 
therefore  be  assumed  to  be  approximately  correct  for  inter- 
mediate cases. 

The  mean  value  of  n  for  iron  and  steel  seems  to  be  \,  so 
that  the  formula  may  be  written 


a'  =  -^  Vt{t  -  y). 


Example. — One  of  the  diagonals  of  a  bowstring  truss  has 
a  sectional  area  of  3  square  inches,  and  is  subjected  to  stresses 


>  \ 


m 


wwW 


m 

• 

Ui 


160 


THEORY  OF  STRUCTURES. 


which  fluctuate  between  a  tension  of  14  tons  and  a  compression 
of  6  tons.     Find  the  statical  strength  of  the  iron. 


a 


„     i4-(- 

-6) 

20 

b    — 

3 

'  3 

d  =  fluctuation  of  stress  = 


t  =  10.17  tons  per  sq.  in. 

9.  Remarks  upon  che  Values  of  t,  u,  s,  and  b. — As  yet 

the  value  of  u  in  compression  '  is  not  been  satisfactorily  detcr- 
nsined,  and  for  the  present  its  value  may  be  assumed  to  be  the 
same  both  in  tens'on  and  compression. 

If,  as  Wohler  states,  "repeated  stresses"  are  detrimental  to 
the  strength  of  a  material,  then  the  values  of  u  and  s  diminish 
as  the  repetitions  increase  in  number,  and  are  minima  in  struc- 
tures designed  for  a  practically  unlimited  life. 

Only  a  very  few  of  Wohler's  experiments  give  the  values  of 
^,  u,  s,  and  a,  so  ihat  Launhardt's  and  Weyrauch's  assumptions 
for  the  value  of /must  be  regarded  as  tentative  only,  and  re- 
quire to  be  verified  by  further  experiments.  The  close  agree- 
ment of  Wohler\>  results  from  tests  upon  untempered  cast-steel 
(Krupp),  with  those  given  by  Launhardt's  formula,  may  be  seen 
from  the  following : 

For  /=  1 100  centners*  per  sq.  zoll,  Wohler  found  that 
u  =  500  centners  per  sq.  zoll.     Thus  (2)  becomes 


«, 


500  I + 


5  ^^, 


). 


and 


.*.  a'  —  500^?,  —  6oo<?j  —  o. 


Hence  for       «,  =    o,       250,     400,     600,     1 100, 
Launhardt's  formula  gives 

rt,  =  500,     710,     800,     900,     1 100; 

*  A  centner  =  110.23  pounds.     A  square  zoll  =  1.0603  square  inches. 


l!-.. 


1. 


REMARKS    UPON    THE   VALUES  OF  t,  U,  S,  AND  b.       l6l 

while  Wohler's  experiments  gave 

«,  =  5C)0,     700,     800,     900,     1 100. 

Again,  with  PhcEnix  iron,  for  /  =  500  centners  per  sq.  zoll, 
u  was  found  to  be  300  centners  per  sq.  zoll,  and 


or 


•.«.  =  300(1+1^;) 


«,  —  300^,  —  250^,  =  o. 


If  rtj  =  240,  a^  =  436.8,  "'hich  almost  exactly  agrees  with 
the  result  given  by  the  tension  experiments. 

In  general,  the  admissible  stress  per  square  unit  of  sectional 
area  may  be  expressed  in  the  form 

(^  ir^v{i  ±  incf>\ (17) 

V  and  in  being  certain  coefifiicients  which  depend  upon  the 
nature  of  the  material  and  also  upon  the  manner  of  the  loading. 
Consider  three  cases,  the  material  in  each  case  being  wrought- 
iron : 

{a)  Let  the  stresses  vary  between  a  maximum  tension  and 
an  equal  maximum  compression  ;  then 


and 


0=1, 

,•.  b  =  700(1  —  i)  =  350*  per  cent.'. 


{l>)  Let  the  material  be  subjected  to  stresses  which  are 
either  tensile  or  compressive,  and  let  it  always  return  to  the 
original  unstrained  condition  ;  then 

min.  B  =  o,     or    max.  B'  =  o,     and     .•.  0  =  0. 

.'.  b  =  700(1  ±  o)  =  700*  per  cent.'. 

{(-)  Let  the  material  be  continually  subjected  to  the  same 
dead  load ;  then 

min.  B  =  max.  B, 


I'  i   '! 


:  i 


1; 


\i 


■f  3: 


h  ; 


Itllf 


■fi 


\i 


i  i ' 


162 

and 


THEORY  OF  STRUCTURES. 


yoo{i  -f-  i)  =  1050*  per  cent.'  =  14,934  lbs.  per  sq.  in., 


which  is  one  third  of  the  ultimate  breaking  strength,  viz.. 
1050*  per  cent.'. 

Thus  in  these  three  cases  the  admissible  stresses  are  in  the 
ratios  of  1:2:3,  ratios  which  have  been  already  adopted  in  ma- 
Chine  construction  as  the  result  of  experience. 

VVohler,  from  his  experiments  upon  untempered  cast-steel 
(Krupp),  concluded  that  for  alternations  between  an  unloaded 
condition  and  either  a  tension  or  a  compression,  b  =  1 100,  and 
for  alternations  between  equal  compressive  and  tensile  stresses, 
d  =  580. 

In  America  it  has  often  been  the  practice  to  take 

max.  B  -\-  max.  B'       a'  -f-  a" 


700 


7CX) 


for  stresses  alternately  tensile  and  compressive,  it  being  as- 
sumed that  if  the  stresses  are  tensile  only,  their  admissible 
values  may  vary  from  o*  to  700*  per  cent.". 


Since  0  z=  —r,  .-.a  = 
a 


700F  a'         700 

,,and.-.^  =  -p  =  ^-^-,.   (US) 


i. 

h 

I 

I, 

560, 

467, 

400, 

350, 

612, 

525. 

437. 

350. 

1  +  0 

Comparing  this  with  (12), 

for  0'  =    o, 
;  (18)  gives  d  —  700, 

and  (12)  gives  fi  =  700, 

10.  Flow  of  Solids. — When  a  ductile  body  is  strained 
beyond  the  elastic  limit,  it  approaches  a  purely  plastic  con- 
dition in  which  a  sufificiently  great  force  will  deform  the  body 
indefinitely.  Under  such  a  force,  the  elasticity  disappears  and 
the  material  is  said  to  be  in  a  ^uu/  state,  behaving  preciscl)- 
like  a  fluid.  For  example,  it  flows  through  orifices  and  shows 
a  contracted  section.  The  stress  developed  in  the  material  is 
called  \.\\Q  flvid  pressure  or  coefficient  of  fluidity. 

The  general  principle  of  the  flow  of  solids,  deduced  by 
Trcsca,  may  be  enunciated  as  follows: 


A  pres 
motion  of 
This  g 
in  materia 
ing,  rolling 
ably  it  als 
certain  ext 
Rails  u 
have  acqui 
due  to  the 
the  rails  an 
of  solids  ar 
and  in  the 
of  fluidity 
disappear  a 
In  punc 
at  first  com 
CDmmences 
brought  un( 
illustration 
I  75  inches, 
length  of  th 
flow  must  h; 
shearing   thi 
really  shorn 
iiuasure   A  t 
area  and  th( 
bserved  by 
-le  fractured 
j,nven  them  a 
'ie  tound  cur 
'•ottom  to  th 
'ating  planes 
and  rolling  th 
In  experi 
lates,  one  ab 
'65,  with  a  h( 
lead  was  alwa 


u  mi 


FLOW  OF  SOLIDS. 


J63 


A  pressure  upon  a  solid  body  creates  a  tendency  to  the  relative 
motion  of  the  particles  in  the  direction  of  least  resistance. 

This  gives  an  explanation  of  the  various  effects  produced 
in  materials  by  the  operations  of  wire-drawing,  punching,  shear- 
ing, rolling,  etc.,  and  in  the  manufacture  of  lead  pipes.  Prob- 
ably it  also  explains  the  anomalous  behavior  of  solids  under 
certain  extreme  conditions. 

Rails  which  have  been  in  use  for  some  time  are  found  to 
have  acquired  an  elongated  lip  at  the  edge.  This  is  doubtless 
due  to  the  flow  of  the  metal  under  the  great  pressures  to  which 
the  rails  are  continually  subjected.  Other  examples  of  the  flow 
of  solids  are  to  be  observed  in  the  contraction  of  stretched  bars 
and  in  the  swelling  of  blocks  under  compression.  The  period 
of  fluidity  is  greater  for  the  more  ductile  materials,  and  may 
disappear  altogether  for  certain  vitreous  and  brittle  substances. 

In  punching  a  piece  of  wrought-iron  or  steel,  the  metal  is 
at  first  compressed  dSidi  flows  inwards,  while  the  shearing  only 
commences  when  the  opposite  surface  begins  to  open.  A  case 
brought  under  the  notice  of  the  author  may  be  mentioned  in 
illustration  of  this.  The  thickness  of  a  cold-punched  nut  was 
1.75  inches,  the  nut-hole  was  .3125  inch  in  diameter,  and  the 
length  of  the  piece  punched  out  was  only  .75  inch.  Thus  the 
flow  must  have  taken  place  through  a  depth  of  I  inch,  and  the 
^lu.iring  through  a  depth  of  .75  inch.  Hence  the  surface 
iLilly  shorn  was  n  X  .3 '25  X  .75  =■  .736  sq.  in.  in  area,  and  a 
niLiisurt  A  the  shearing  action  is  the  product  of  this  surface 
area  and  the  fluid  pressure.  The  nature  of  the  flow  may  be 
bserved  by  splitting  a  cold-punched  nut  in  half  and  treating 
iiu'  fractured  surfaces  with  acid,  after  having  planed  them  and 
Ljivrn  them  a  bright  polish.  The  metal  bordering  the  core  will 
i)c  lound  curved  downwards,  the  curvature  increasing  from  the 
Dottom  to  the  top,  and  well-defined  curves  will  mark  the  sepa- 
rating planes  of  the  plates  which  were  originally  used  in  piling 
and  rolling  the  iron. 

In  experimenting  upon  l-^ad.  Tresca  placed  a  number  of 
>lates,  one  above  the  other,  in  a  -trong  cylinder.  Fig.  188,  page 
i6>,  with  a  hole  in  the  bottom.  Upon  applying  pressure  the 
lead  was  always  f uund  to  flow  when  the  coefficient  of  fiuidity 


m 


\mm 


■  1 1 


164 


THEORY  OF  STRUCTURES. 


I 


was  about  2844  lbs.  per  sq.  in.,  the  difference  of  stress  beini; 
double  this  amount.  The  separating  planes  assumed  curved 
forms  analogous  to  the  corresponding  surfaces  of  flow  when 
water  is  substituted  in  the  cylinder  for  the  lead. 

The  flow  of  ductile  metals,  e.g.,  copper,  lead,  wrought-iron, 
and  soft  steel,  commences  as  soon  as  the  elastic  limit  is  ex- 
ceeded, and  in  order  that  the  flow  may  be  continuous  the  dis- 
torting stress  must  constantly  increase.  On  the  other  hand, 
in  the  case  of  truly  plastic  bodies,  flow  commences  and  con- 
tinues under  the  same  constant  stress.  It  evidently  depends 
upon  the  hardness  of  the  material,  and  has  been  called  the  co- 
efficient of  hardness.  The /(3;/^rr  the  stress  acts  the  greater  is 
the  deformation,  which  gradually  increases  indefinitely  or  at  .1 
diminishing  rate. 

Experiment  shows  that  there  is  very  little  alteration  in  the 
density  of  a  ductile  body  during  its  plastic  deformation,  anil 
Tresca's  analytical  investigations  are  based  on  the  assumption 
that  the  body  is  deformed  without  sensible  change  of  volume. 

Consider  a  prismatic  bar  undergoing  plastic  deformation. 

Let  L  be  the  length  and  A  the  section  of  the  bar  at  com- 
mencement of  deformation. 

Let  Z  -|-  .r  be  the  length  and  a  the  section  of  the  bar  at  a 
subsequent  period. 

Let/  be  the  intensity  of  the  fluid  pressure. 

Since  the  volume  remains  unchanged, 

LA={L±  x)a, (i) 

the  positive  or  negative  sign  being  taken  according  as  the  bar 
is  in  tension  or  compression. 

Let  /^,  be  initial  force  on  bar. 

Let  P  be  force  on  bar  when  its  length  \s  L  ±.  x.    Then 


and  the  fc 


-L- 


y 


CD  of  a  cyl 
D.  A  hole 
oi  the  fac( 
flows  under 
by  a  piston, 
pressed  to 
the  correspc 

First,  as 
the  mass  rei 

If  dx  be 
DO  correspc 
length  of  th 


Integrati 
.)'  =  o  when 


P.=pA. 

P=pa, 

and  hence 

Pa           L 

P,  ~  A"  L±x 

.     .     (2) 

Hence 

P{L  ±  x)  =  P,L  =  a  constant,  .     . 

.     .     (3) 

Second,  a 
(d/r  transfer 
'-^  L'N'lindrical 


FLOW   OF   SOLIDS. 


165 


V 


and  the  force  diminishes  a  >  the  bar  stretches  and  increases  as 

the  bar  contracts  under  pressure. 
If  equation  (3)  be  referred  to  rect- 
angular axes,  the  ordinates  repre- 
senting different  values  of  P  and 
the  abscissc-E  the  corresponding 
values  of  x,  the  stress-strain  dia- 
'■%  grams,  tt  in  tension  and  cc  in  corri- 
pression,  are  hyperbolic  curves, 
having  as  asymptotes  the  axis  of 
X,  XOX,  and  a  line  parallel  to  the 
axis  of  _;v  at  a  distance  from  it 
equal  to  the  length  L  of  the  bar. 

Next  consider  a  metallic  mass 
(e.g.,  lead)  resting  upon  the  end 

CD  of  a  cylinder  of  radius  R,  and  filling  up  a  space  of  depth 

D.     A  hole  of  radius  r  is  made  at  the  centre 

of  the   face  CD,  through  which   the   mass 

flows  under  the  pressure  of  fluidity  exerted 

by  a  piston.     When  the  mass  has  been  com- 
pressed to  the  thickness  DO  =■  x,  let  y  be 

the  corresponding  length  KE  of  the  "jet." 
First,  assume  that  the  specific  weight  of 

the  mass  remains  constant. 

If  dx  be  the  diminution  in  the  thickness 

DO  corresponding  to  an  increase  dy  in  the 

length  of  the  jet,  then 


Fig.  187. 


nICdx  -\-  Ttr^dy  =  0. 


(0 


Integrating  eq.  I,  and  remembering  that 
1'  =  o  when  X  =.  D, 

R\D  —  x)-  ry  ^  o. 


(2) 


Second,  assume  that  the  cylindrical  portion  EFGH  \%  gradti- 
ally  transformed  into  NMPLKQN,  of  which  the  part  PMNQ 
is  cylindrical,  while  the  diameter  of  the  part  PLKQ  gradually 


otic 

too 


THEORY  OF  STRUCTURES. 


increases  from  the  face  of  the  cylinder  to  KL  (  =  EF\  at  tlie 
end  of  the  jet.    Then 


n{R*-r')dx 


amount   of   meta!   which   flows   into   the 
central  cylinder 


=  27trdrx, 


(3) 


dr  being  the  depth  to  which  the  metal  penetrates. 

Third,  assume  that  the  diminution  of  the  diameter  of  the 
cylindrical  portion  PMNQ  is  directly  proportional  to  the  said 
diameter. 

Then,  if  2  be  the  radius  of  the  cylinder  PQNM, 


dr       dz 
r  ~  z  ' 


By  eqs.  (3)  and  (4), 


X  2 


Integrating, 


{R'-r')\og,x  =  2r'\og,z-{-c, 


c  being  constant  of  integration. 
When  X  =  D,  z  —  r, 


.'.  (^"-01og,-^-  =  2rMog,|, 


(4) 


or 


/f»-r» 


2  _  lx\    ='■' 
7~  \DI 


(5) 


I     1 


WORK. 


By  eqs,  (2)  and  (5), 


^' 


187 


(6) 


which  is  the  equation  to  the  profile  PL  or  QK. 

Note. — If  K'  =  sr",  eq,  (6)  represents  a  straight  line. 
''  F^  =  2r\       "  "  "  parabola. 

II.  Work. — Work  must  be  done  to  overcome  a  resistance. 
Thus  bodies,  or  systems  of  bodies,  which  have  their  parts  suit- 
ably arranged  to  overcome  resistances  are  capable  of  doing 
work  and  are  said  to  possess  energy.  This  energy  is  termed 
kinetic  ox  potential  according  as  it  is  due  to  motion  or  to  posi- 
tion. A  pile-driver  falling  from  a  height  upon  the  head  of  a 
pile  drives  the  pile  into  the  soil,  doing  work  in  virtue  of  its 
motion.  Examples  of  potential  energy,  or  energy  at  rest,  are 
afforded  by  a  bent  spring,  which  does  work  when  allowed  to 
resume  its  natural  form  ;  a  raised  weight,  which  can  do  work  by 
falling  to  a  lower  level ;  gunpowder  and  dynamite,  which  do 
work  by  exploding ;  a  Leydcnjar  charged  with  electricity,  which 
does  work  by  being  discharged ;  coal,  storage  batteries,  a  head 
of  water,  etc.  It  is  also  evident  that  this  potential  energy 
must  be  converted  into  kinetic  energy  before  work  can  be 
done.  A  familiar  example  of  this  transformation  may  be  seen 
in  the  action  of  a  common  pendulum.  At  the  end  of  the 
swing  it  is  at  rest  for  a  moment  and  all  its  energy  is  potential. 
When,  under  the  action  of  gravity,  it  has  reached  the  lowest 
point,  it  can  do  no  more  work  in  virtue  of  its  position.  It  has 
acquired,  however,  a  certain  velocity,  and  in  virtue  of  this 
velocity  it  does  work  which  enables  it  to  rise  on  the  other  side 
of  the  swing.  At  intermediate  points  its  energy  is  partly 
kinetic  and  partly  potential. 

A  measure  of  energy,  or  of  the  capacity  for  doing  work,  is 
the  zvork  done. 

The  energy  is  exactly  equivalent  to  the  actual  work  done 
in  the  following  cases  : 

{a)  If  the  effort  exerted  and  the  resistance  have  a  common 
point  of  application. 


H 


If 


If'' 


ill! 


Iffi^ 


i 


!  i 
; 


l\ 


\i 


r: 


li 


i68 


YHEORY  OF  STRUCTURES. 


{b)  If  the  points  of  application  are  different  but  are  rigidly 
connected. 

{c)  If  the  energy  is  transmitted  from  member  to  member, 
provided  the  members  do  not  change  form  under  stress,  and 
that  no  energy  is  absorbed  by  frictional  resistance  or  restraint 
at  the  connections. 

Generally  speaking,  work  is  of  two  kinds,  viz.,  internal  work, 
or  work  done  against  the  mutual  forces  exerted  between  the 
molecules  of  a  body  or  system  of  bodies,  and  external  zvork,  or 
work  done  by  or  against  the  external  forces  to  which  the  bod\- 
or  bodies  are  subjected.  In  cases  {a),  {b),  {c)  above,  the  inter- 
nal work  is  necessarily  nil. 

As  a  matter  of  fact,  every  body  yields  to  some  extent  under 
stress,  and  work  must  be  done  to  produce  the  deformation. 
Frictional  resistances  tend  to  oppose  the  relative  motions  of 
members  and  must  also  absorb  energy.  If,  however,  the  work 
of  deformation  and  the  work  absorbed  by  frictional  resistance 
are  included  in  the  term  work  done,  the  relation  still  holds  that 

Energy  =  work  done. 

A  measure  of  work  done  is  the  product  of  the  resistance  by 
the  distance  through  which  it  is  overcome.  When  a  man 
raises  a  weight  of  one  pound  one  foot  against  the  action  of 
gravity  he  does  a  certain  amount  of  work.  To  raise  it  two  feet 
he  must  do  twice  as  much  work,  and  ten  times  as  much  to  raise 
it  ten  feet.  The  amount  of  work  must  therefore  be  propor- 
tional to  the  number  oi  feet  through  which  the  weight  is 
raised.  Again,  to  raise  two  pounds  one  foot  requires  twice  as 
much  work  as  to  raise  one  pound  through  the  same  distance ; 
while  five  times  as  much  work  would  be  required  to  raise  five 
pounds,  and  ten  times  as  much  to  raise  ten  pounds.  Thus  the 
amount  of  work  must  also  be  proportional  to  the  weight  raised. 
Hence  a  measure  of  the  work  done  is  the  product  of  the 
number  of  pounds  by  the  number  of  feet  through  which  they 
are  raised,  the  resulting  number  being  designated  foot-pounds. 
Any  other  units,  e.g.,  a  pound  and  an  inch,  a  ton  and  an 
inch,  a  kilogramme  and  a  metre,  etc.,  may  be  chosen,  and  the 
work   done   represented  in  inch-pounds,  inch-tons,  kilogram- 


=  A'  cos  6  arc 


\n. 


OBLIQUE  RESISTANCE. 


169 


metres,  etc.  This  standard  of  measurement  is  applicable  to  all 
classes  of  machinery,  since  every  machine  might  be  worked  by 
means  of  a  pulley  driven  by  a  falling  weight. 

12.  Oblique  Resistance. — Let  a  body  move  against  a 
resistance  R  inclined  at  an  angle  0  to  the  direction  of  motion 
|Fig.  189).  No  work  is  done  against  the 
normal  component  R  sin  0,  as  there  is  ^~ 
no  movement  of  the  point  of  applica-  \ 
tion  at  right  angles  to  the  direction  of 
motion.  This  component  is,  there- 
fore, merely  a  pressure.  The  work 
done  against  the  tangential  component 
R .  cos    8    between    two    consecutive  ^"^'  '^' 

points  M  and  A''  of  the  path  of  the  body  is  R  cos  B .  MN. 
Hence  the  total  work  done  between  any  two  points  A  and  B  of 
the  path 

=  2{R  cos  e .  MN)  =  f'R  cos  dds, 

s  being  the  length  of  AB. 

If  AB  is  a  straight  line  (Fig.  190),  and  if  R  is  constant  in 
direction  and  magnitude, 

the  total  work  =  R  cos  0 . AB  =  R.AC, 

.-iC"  being  the  projection  of  the  displacement  upon  the  line  of 
action  of  the  resistance.     Let  the  path  be  the  arc  of  a  circle 

Rsiae   R 


Fig.  190.  Fig,  191. 

iFig.  191)  subtending  an  angle  a  at  the  centre.     If  R  and  6  re- 
main  constant,  the  work  done  from  A  to  B 

=  K  cos  e  arc  AB  =  R  cos  0  .  OA  .  a  =  R  .  OM cosB.a 

=  Rpa  =  Ma, 


■i 


Is   1 


w 


II  ll     ■■■:    '>■  ;    J 


"•!f" 


m 


I? 


M 


fjllli 


■-'I';' 


IMAGE  EVALUATION 
TEST  TARGET  (MT-3) 


1.0 


I.I 


-  m 


22 

1.8 


1.25      1.4 

1.6 

-m 6"     — 

► 

Photographic 

Sciences 

Corporation 


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1^         ^"^'m 


f/j 


«^ 


m 


170 


THEORY  OF  STRUCTURES. 


p  being  the  perpendicular  from  O  upon  the  direction  of  /?,  and 
M  =  Rp  being  the  moment  of  resistance  to  rotation. 

If  there  are  more  resistances  than  one,  they  may  be  treated 
separately  and  their  several  effects  superposed.  In  such  case, 
M  will  be  the  total  moment  of  resistance  and  will  be  equal  to 
the  algebraic  sum  of  the  separate  moments. 

The  normal  component  R  sin  6  produces  a  pressure. 

13.  Graphical  Method. — Let  a  body  describe  a  path  AB 

B  (Fig.  I92)against  a  variable  resistance  of 
such  a  character  that  its  magnitude  in 
the  direction  of  motion  may  be  repre- 
sented at  any  point  M  by  an  ordinate 
MN  to  the  curve  CD.  Let  the  path 
AB  be  subdivided  into  a  number  of 
parts,  each  part  MP  being  so  small 
that  the  resistance  from  M  to  P  may 
be   considered   uniform.      The   mean 

r    ,.         .                 MN-^PQ        J    ^ 
value  of  this  resistance  = ,  and  the  work  done  in 

2 

MN-\-PQ 
overcoming  it  = .  MP  —  the  area  MNQP  in  the 

limit.  Hence  the  total  work  done  from  A  to  B  —  the  area 
bounded  by  the  curves  AB,  CD  and  the  ordinatcs  AC,  BD. 

14.  Kinetic  Energy. — The  velocity  v  acquired  by  a  body 
of  weight  tt>  and  mass  m  in  falling  freely  from  rest  through  the 
vertical  distance  //  is 

V  =  ^2gh ; 

,         W  V  V* 

.*.  wli  = =  m—, 

g2  2 

Thus  an  amount  of  work  zvh  is  done,  and  the  body  possesses 

the  kinetic  energy  ;;/— . 

Again,  let  v'  be  the  velocity  of  the  body  after  falling 
through  a  further  distance  x,  measured  vertically.    Then 


w{h  -f-  ^)  = 


;//?' 


and 


Ktl^Eftt  EMEkdV. 


.*.  wx  =  --\v    —  tr). 


171 


Thus  the  work  done  in  falling  through  the  vertical  distance  x 
is  zvx,  and  is  equal  to  the  corresponding  change  of  kinetic 
energy. 

15.  Example  i.  Let  it  be  required  to  determine  the  7uork 
done  in  stretching  or  compressing  a  bar  of  length  L  and  sec- 
tional area  A  by  an  amount  /. 

Suppose  that  the  force  applied  to  ihe  bar  gradually  in- 
creases from  o  until  it  attains  the  value  /*;  its  mean  value  is 

P  P 

-,  and  the  work  done  is  therefore  -/. 

2  2  , 

/ 
But  P  =  EAj  ;  E  being  the  coefTicient  of  elasticity. 


■i 


.  h 


% 


4     ';■,«!? 


.*.  the  work  done 


2      L 


J  IPVAL 
E\AI     2  ' 


This  formula  is  only  true  for  small  values  of  the  ratio  j. 

In  the  case  of  a  compressive  force  it  is  assumed  that  the  bar 
does  not  bend. 

P      . 

A  suddenly  applied  force,  -  ,  will  do  as  much  work  as  a 

steady  force  which  increases  uniformly  from  o  to  /",  and  hence 

it  follows  that  a  bar  requires  twice  the  strength  to  resist  with 

safety  the  sudden  application  of  a  given  load  than  is  necessary 

when  the  same  load  is  gradually  applied. 

If /is  tho  proof  stress  or  elastic  limit  per  unit  of  sectional 

/ 
area, -A  is  the  corresponding /rtf<?/"j/rrti«,  and  the  work  done  in 
E 

producing  the  latter  is  called  the  resilience  of  the  bar.     Accord- 

f*  AL    /' 
ing  to  the  above,  its  value  is  -^  —  ;  -p   is  called  the  Modulus 

of  Resilience. 


m 

Vt'.'  ■  t 


:.:i-i. 


'^'ih. 


173 


THEORY  OF  STRUCTURES. 


Ex.  2.  A  wrought-iron  tie-rod,  30  ft.  in  length  and  4  sq. 
in.  in  sectional  area,  is  subjected  to  a  longitudinal  pull  of 
40,000  lbs.  Determine  the  unit  stress,  the  strain,  and  the  elon- 
gation, the  coefficient  of  elasticity  being  30,000,000  lbs. 

40000 
The  unit  stress  is  — -^ —  =  10,000  lbs.  per  sq.  in. 

Also,  from  the  elastic  law,  loooo  =  30000000  X  strain. 

.*.  the  strain  =  p^ 
and  the  elongation  =  lis  =  IS)  ft. 

Ex.  3.  A  steel  rod  is  15  ft.  long  and  2^  sq.  in.  in  sec- 
tional area.  The  proof  strain  of  the  steel  is  j^,  and  its  coeffi- 
cient of  elasticity  is  36,000,000  lbs.  Find  the  greatest  weight 
that  can  be  safely  allowed  to  fall  upon  the  end  of  the  rod  from 
&  height  of  27  ft. 

The  proof  stress  =  £'X  proof  strain  =  36,000  lbs.  persq.  in. 

The  compression  of  the  rod  under  the  proof-stress  is 

—  =  -  f  t 
1000  —  300  "■• 

The  resilience  of  the  rod 

_r  AL  _  (36000)'  2i  X  15  X  12 

~  E    2    ~  36000000  2 

=  8100  inch-lbs.  =  675  ft.-lbs.  * 

.     Again,  let  Whe  the  required  weight  in  pounds. 

The  total  distance  through  which  it  falls  =  27  ft.  -f-  com- 
pression =  (27  -f-  Js)   feet,  and   the    corresponding    work    is 

W{2y  -f-  ^)  ft.-lbs.     This  must  of  course  be  exactly  equivalent 
to  the  resilience  of  the  rod,  and 


and 


W  —  24.9  lbs. 


KINETIC  ENERGY. 


173 


The  resilience  of  the  rod  may  also  be  at  once  found  from 
the  fact  that  it  is  the  product  of  one  half  of  the  total  stress  by 
the  compression,  i.e.,  ^  .2\.  36000  X  i!^  =  675  ft.-lbs. 

Ex.  4.  Let  w^ ,  «', ,  «', ,  .  .  .  7c„  be  the  weights  of  a  system 
of  particles  rigidly  connected  together  and  at  distances  x^,  x^, 
x\,  .  .  .  x„,  respectively,  from  a  given  axis.  Let  the  system 
revolve  around  the  axis  with  a  uniform  angular  velocity  A. 

The  kinetic  energies  of  the  several  particles  are 


«',  x.^A''       n\  x^A* 


w,  x*A* 


g      2    '      g       2     '  '      g      2    ' 

and  therefore  the  total  kinet'c  energy  of  the  system 


=  ^--  -]  w,x,^  -\-  ^,.1-,"  4-  .  .  .  -f  Zi'^xJ'  [ 


2g  { 


"  s 


A''  (  ) 

=    —  -i  '«,'f  1'  +  Wj-l'a'  +  •  •  •  +  W«'^h'  [ » 

w, ,  >n,  ,  .  .  .  iu„  being  the  masses  of  the  particles. 

The  sum  between  the  brackets  is  called  the  moment  of  in- 
ertia of  the  system  of  particles  about  the  axis  and  is  usually 
denoted  by  /. 


.*.  the  total  kinetic  energy  = 


A' I 


'         I 


ii 

b.  ,1 


Again,  it  appears  from  the  definition  that  every  moment  of 
inertia  is  the  product  of  a  mass  and  the  square  of  a  length. 
This  length  is  called  the  radius  of  gyration  and  is  usually 
designated  by  the  symbol  k. 

If  M  be  the  total  mass  of  the  system,  and  W  the  total 
weight, 

W 
S 

,',.      .  i^Mf^y       W  {Akf     , 

and  the   total   kmetic   energy  =  M = ~.  the  re- 

&/  2  g      "2- 


,^ 


M  r     S 


r  ? 


>74, 


THEORY  Of  structures. 


suit  being  the  same  as  if  the  particles  were  collected  in  a  ring 
of  radius  k,  sometimes  called  the  equivalent  ring  or  fly-wheel. 
Let  Ig  be  the  moment  of  inertia  of  the  system  with  respect 
to  a  parallel  axis  through  the  centre  of  gravity,  and  let  //  be  the 
distance  between  the  two  axes.     Then 

I,  =  mih  -  x,Y  +  m,{h  -x,Y +...-]-  mlh  -  ^„)» 
=  ie'2(in)  —  2h'2{mx)  +  2(w4r'). 

Since  the  new  axis  passes  through  the  centre  of  gravity, 

"^mx  =  MA. 

Also,  2{m)  =  M    and     2{mx')  =  /; 

.-.  /^  =  M/i'  4-  /  -  2M/1' ; 
.-.  I^I,-\-Mk\ 

So,  if  /'  is  the  moment  of  inertia  about  another  parallel  axis 
at  the  distance  //'  from  the  centre  of  gravity. 


I'  =I,-\-Mir 

.\i-Mw  =  r  ■ 


Mh'\ 


i  i    , 


iil' 


Hence,  if  the  positions  of  two  parallel  axes  relatively  to  the 
centre  of  gravity  are  known,  and  if  the  moment  of  inertia  about 
one  is  given,  the  moment  of  inertia  about  the  other  can  be 
obtained  by  means  of  the  last  formula. 

Note. — Nothing  has  been  said  as  to  the  number  of  the  par- 
ticles. They  may  be  infinite  in  number  and  infinitely  near 
each  other,  forming  in  fact  a  solid  body.  The  summation 
2{mx')  is  then  best  effected  by  integration. 

16.  Values  of  A;*. 

1.  For  a  rectangular  plate  of  depth  d  with  re- 

spect  to  an  axis  through   the  centre  _ 

perpendicular  to  the  side  d. ^'  =  — . 

2.  For  a  circular  plate  of  radius  r  with  re-  , 

spect  to  a  diameter k*  =  —, 


VALUES  OF  k\ 


175 


3.  For  an  annulus  of  external  radius  r,  and 
internal  radius  r,  with  respect  to  a 
diameter ^' 

Note. — If  r,— r,  =  t,  and  the  breadth 
t  of  the  annulus  is  small  as  compared 
with  the  radius  r, ,  then 


_  K  4-  r: 


k'  —    '   '  ^  ' — '-  —  -^^ — '-,  approx., 
4  2 


and  the  area 


=  n{r^  —  r/)  =  2nr^t,  approx. 


4.  For  the  plates  in  (2)  and  (3)  with  respect 
to  an  axis  through  the  centre  perpen- 
dicular to  the  plates,  the  numerators 
remain  the  same  but  the  denominator 
is  in  each  case  2. 

'5.  For  a  sphere  of  radius  r  with  respect  to  a 

diametCi k^ 

6.  For  a  solid  cylinder  of  radius  r  with  re- 

spect to  its  axis k^ 

7.  For  an  elliptic  plate  of  which  the  major 

and  minor  axes  are  2b  and  id  respec- 
tively : 

With  respect  to  the  major  axis /•' 

With  respect  to  the  minor  axis k^ 

8.  For  a  triangular  plate  of  height  //  with  re- 

spect  to  an   axis  coinciding  with  the 
base .   k^ 


5 

r' 

2 


4 
b' 


1^ 
6" 


.ill    ' 


'r    I 


\m 


\W\ 


1" 


Tll^ 


• ' 


I 
i 


IH.  >■ 


lj6 


THEORY  OF  STRUCTURES. 


17.  Momentum— Impulse. — A  moving  body  of  weight  it' 
and  mass  m  acted  upon  in  the  direction  of  motion  for  a  time  / 
by  a  force  F  will  acquire  a  velocity  v  which  is  directly  propor- 
tional to  /^and  to  t,  and  inversely  proportional  to  w.     Hence 


V  —  n- 


Ft 


w 


n  being  some  coefificient. 

If  F=  tf,  the  velocity  generated  in  one  second  is^. 


and 


or 


•••  g=n. 


Ft       Ft 
.'.  v=g—  =  —  , 


mv  =  Ft. 


This  is  the  analytical  statement  of  Newton's  Second  Law 
of  Motion,  which  has  been  expressed  by  Clerk  Maxwell  in  the 
following  form  :  "  The  change  of  momentum  (i.e.,  the  product  of 
the  mass  and  velocity)  is  numerically  equal  to  the  impulse  (i.e., 
the  product  of  the  force  and  the  time  during  which  it  acts) 
xvhicU  produces  it,  and  is  in  the  same  direction." 

Again,  let  p  be  the  perpendicular  from  a  fixed  axis  O  upon 
the  direction  of  motion  of  the  body,  and  let  r  be  the  radius  OP 
to  the  body.    Then 

imp  =  Ftp  —  Fpt  =  Mt, 

where  M  =  Fp\  or  the  change  of  the  moment  of  momentutn,  i.e., 
of  the  angular  momentum,  is  equal  to  the  moment  of  impulse. 

The  above  results  are  also  true  for  two  or  more  bodies  or 
systems  of  bodies  severally  acted  upon  by  extraneous  forces, 
and  the  equations  may  be  written 

2mv  =  2Ft,     2mvp  =  2Fpt  =  2Mt. 

In  words,  the  total  change  of  momentum  in  any  assigned  direction 
is  equal  to  the  algebraic  sum  of  the  impulses  in  the  same  direction. 


MOMENTUM— IMP  ULSE. 


177 


/'. 


y^\ 


y 


T 


and  the  total  change  of  angular  momentum  is  equal  to  the  alge- 
braic sum  of  the  moments  of  the  impulses. 

Hence  it  follows  that  if  two  or  more  bodies  or  systems  of 
bodies  mutually  attract  or  repel  each  other,  and  if  there  are  no 
extraneous  forces,  the  total  momentum  in      «,  yy 

any  assigned  direction  is  constant  (the 
principle  of  the  conservation  of  linear 
momentum),  and  the  angular  momentum         /  x 

about  a  given  axis  is  constant  (the  prin- 
ciple of  the  conservation  of  angular 
momentum).  /,''' 

Suppose  that  the  velocity  of  the  body  0*^':" r^. 

of  weight  zv  and  mass  m  changes  from  "•^v„  / 

t',  to  %\  in  the  time  /  under  the  action  of  "\  /' 

a  couple  of  moment  M,  and  let/,,/,  be  F'g-  '93- 

the  corresponding  values  of  p,  and  r, ,  r,  those  of  r,  Fig.  193. 

.-.  ;«(t'j>,  -  v^^  =  Mt-  ' 

or  if  w, ,  w^  are  the  components  of  i\,  f,  in  directions  perpen- 
dicular to  r, ,  r,,  respectively, 

m{u\r^  —  ii\r^  =  Mt. 

For  example,  a  weight  W  of  water  passing  through  a  turbine 

of  external  radius  r,  and  internal  radius  r,  has  its  angular  mo- 

W  W 

mentum    changed   from   — w,r,    to   — 't^'3^3,    w, ,    ^v.^    being 

the  tangential   components   of   the  velocity  with  which   the 

water  enters  and  leaves  the  wheel.    The  water,  therefore,  exerts 

W 
upon  the  wheel  a  couple  of  moment  —  (tf,r,  —  w,rj,  and    if 

the  wheel  rotates  with  an  angular  velocity  A,  the  work  done 
upon  the  wheel  by  the  water 

W  W 

=  --^(ic'.r,  -  w^r^  =  —  (w,«,  -  w,u,), 

u,  and  «,  being  the  circumferential  velocities  corresponding  to 
?•,  and  r,,  respectively. 


m 


.,v 


•'  'viF 


viiliy' 


'♦f'' 


if't 


i 


m 


I     I 


I 


178 


THEORY  OF  STRUCTURES. 


I 


i8.  Useful  Work— Waste  Work. — Let  a  body  of  mass  in 
and  weight  w  pass  over  the  distance  s  under  the  action  of  a 
force  F  acting  in  the  direction  of  motion  for  a  time  /,  and  let 
the  velocity  of  the  body  change  from  ?',  to  7',.  Assume  /  to 
be  so  small  that,  for  the  interval  in  question,  the  velocity  may 

be  regarded  as  constant  and  of  the  average  value  -^— - — - ; 


s  = 


V,  +  V, 


t. 


But  Ft  =  {mv^  —  mv^. 


m 


Fts=^-{v:-v,y, 


or 


tn 


Fs  =  -iv:  -  v,% 


t 


Thus  Fs,  the  work  done,  is  equal  to  the  change  of  kinetic 
energy  in  the  given  interval. 

If  the  body  is  a  material  particle  of  a  connected  system,  a 
similar  relation  hcilds  for  every  other  particle  of  the  system,  and 
the  total  work  done  =  \{'2mv.^  —  ^viv^). 

A  part  of  thio  work  may  be  expended  in  doing  what  is 
called  effective  work,  i.e.,  in  overcoming  (i)  an  external  resist- 
ance, or  in  doing  useful  work,  and  (2)  frictional  resistance,  or  in 
doing  xvasted  work. 

Denoting  the  total  effective  work  by  T,  and  the  total  motive 
work  by  T^ ,  tne  last  equation  may  be  written 

7;-  T,  =  \{'2mv:-:2mv^\ 

and  the  difference  between  the  total  motive  work  and  the  total 
effective  work  is  equal  to  the  total  change  of  kinetic  energy. 

In  the  case  of  a  machine  working  at  a  normal  speed  the 
velocities  of  the  different  parts  are  periodic,  being  the  same  at 
the  beginning  and  end  of  any  period  or  number  of  periods. 
For  any  such  interval,  therefore,  v^  =  v,,  and  .•.  T„  =  T,,  so 


GENERAL   CASE. 


179 


that  there  is  an  equality  between  the  motive  work  and  the 
effective  work. 

19.  General  Case.— Let  x,,  ^, ,  x;,  be  the  co-ordinates  of 
tlie  C.  of  G.  of  a  moving  body  of  mass  A/ with  respect  to  three 
itctangular  axes  at  any  given  instant. 

Let  x\,  >', ,  xr,  be  the  co-ordinates  of  ilic  same  point  after  a, 
unit  of  time. 

Let  ^, ,  J, ,  ^,  be  the  co-ordinates  of  any  particle  of  mass  m 
at  the  given  instant. 

Let  ^, ,  j,j,  ^,  be  the  co-ordinates  of  the  sdme  particle  after 
a  unit  of  time. 

.-.  Mx,  =  ^Xw;jr,),     M^,  =  ^{my^),     M7,  =  2{ms,) ; 
Mx,  =  2{mx,),     Mj>^  =  2{>'iy^,    M'z^  =  2(»ic,)', 
•••  ^(-i-,  -  ^'.)  =  2w{x,  -  .r,),     M{j',  -  J,)  =  2m{}',  -  7,), 


or 


Mu  =^  2»iu,     Mv  =  2f>iv,    Mw  =  2mw, 


iiy  V,  w  being  the  component  velocities  of  the  C.  of  G.  at  the 
given  instant  with  respect  to  the  three  axes,  and  u,  v,  w  the 
component  velocities  of  the  particle  ;«  at  the  same  instant. 
From  these  last  equations, 

Mil  =  2muu,    Mv  =  2mz'v,    Mza  =  2mww. 

.'.  M{h  -\-  V  -\-w)=-  2m{mi  -\-vv-{-  ivw), 

which  may  be  written  in  the  form 

M(u  +V  4-  ^')  +  ^>«K«  -  "^y  -\-{v-  vY  +  (w  -  wf\ 
MU'  +  2m  V  =  2viv\ 


or 


!•! 


'.i'» 


n 


i; 


,3 


.'I 

m 

•■A 


I 


- 


.  If--; 


til 


i8o 


TliKOKV  or  SlKUCrVh'ES 


V  hv'\\\\^  tho  resultant  vcKnity  of  the  (.'.  of  li. ;  ;•,  that  of  tl\c 
parliclr  ;  antl  /',  that  of  tlu>  paiiide  ichitivcly  to  the  C.  of  ll. 
The  hist  equation  may  he  written 

J/T'       i';///"       i'w;-' 


Thus  the  energy  of  tiie  total  mass  eolleeteil  at  the  centre 
oi  j^ravity,  toj^olher  with  the  enei^jy  relatively  to  the  ceiitn-  of 
j;ravity.  is  equal  to  the  total  enerj^y  of  motion. 

If  the  botly  revolves  arouml  an  axis  throu[;h  its  C.  of  G. 
with  an  angular  veloeitj-  ./.  the  seeoml  term  of  the  last  e(|ua- 
tioi\  becomes 


~:£mr\r 


V  A* 

-    2.titr  ~  —  /, 

2  2       ' 


Total  energy  of  explosion  =  energ^y  of  shot  -|-  energy  of  recoil ; 
Energy  of  shot  =  energy  of  translation  -|-  energy  of   rotation 

_  50     (3020V       500    I    f!^-li   Jo_2oy /3i535\' 
~  32.-  '       2      "^  32:2  ■  2  •  V  -iV  '  ■  AO.\%i   \    12   I 

=  31680124.2+97758.6 

=  31777882.S  ft.-lbs. ; 


.,      36  X  2240    (16JV  ,         ,    ,^ 

Emrgy  of  ncoil  =  ^~^:;^~    .  — ,—  =  330652.1  ft.-lbs. 


0"~ 


'  m 


CENTKIFUGAI.    lOS'CE.  l8l 

ilciicc,  if  Cbe  the  energy  of  i  lb.  of  powder, 

'  C.  300  =  3i777«H2.8  4-  330652.1 
=  32108534.9  ft.-lbs., 
iiul  hence  ^ 

C=  107028.45  ft.-lbs  —  47.7  ft.-tons. 

Ex.  2.  Let  W^be  the  weight  of  a  fly-wheel  in  lbs.,  .ind  let 
its  max.  antl  min.  anj^ular  velocities  be  //,,//,,  respectively. 
The  motion  beinj;  one  of  rotation  only,  the  enerj^y  stored  up 
when  the  velocity  rises  from  //,  to  .',,  or  ^ivin  out  when  it 
falls  from  /J,  to  /i,,  is 

1  IV  IV 

-^  {a;  -  a:)  =  -^^n^i:  -  a,-)  =  ^^;  {v;  -  v^), 


il 


PI 


*    ;.     "11 


'^\ 


1 


t', ,  ?',  being  the  linear  velocities  corresponding  to  A,,  /!,,  and 
I'  being  taken  equal  to  the  mean  radius  of  the  wheel. 

It  is  usual  to  specify  that  the  variation  of  velocity  is  not 
to  exceed  a  certain  fractional  part  of  the  mean  velocity. 

Let  V  be  the  mean  velocity,  and  -  the  fraction.    Then 


V 
V,  —  t\  =  -- ;    also    v,-{-v,~2V; 


v;  -  v:       V* 


w  V^ 
Hence  the  work  stored  or  s;tven  out  = , 


21.  Centrifugal  Force. — A  body  constrained  to  move  in 
a  plane  curve  exerts  upon  the  body  which  constrains  it,  a  force 


\1  ~  .   i 


182 


THEORY  OF  STRUCTURES. 


•0 
Pig.  194. 


called  centrifugal  force,  which  is  equal  and  opposite  to  the  de- 
viating (or  centripetal^  force  exerted  by 
the  constraining  body  upon  the  revolving 
body. 

Let  a  particle  of  mass  m  move  from  a 
point  Z'  to  a  consecutive  point  Q  (Fig.  1 94) 
of  its  path  during  an  interval  of  time  t 
under  the  action  of  a  normal  deviating 
force. 

Let  the  normals  at  P  and  Q  meet  in  O ;  PQ  may  be  con- 
sidered as  the  indefinitely  small  arc  of  a  circle  with  its  centre 
at  0. 

If  there  were  no  constraining  force,  the  body  would  move 
along  the  tangent  at  Z'  to  a  point  T  ^uch  that  PT  =  vt,  v 
being  the  linear  velocity  at  P. 

Under  the  deviating  force  the  body  is  pulled  towards  O 
through  a  distance  /W=  ^/V',/ being  the  normal  accelera- 
tion, and  QN  being  drawn  perpendicular  to  OP, 
Also,  in  the  limit,     PQ  =  PT  =  QN  =  vt. 


But 


QN'  =  PN.  2OP', 
.'.  v'f  -  \ff2R, 


R  being  the  radius  0P\  and  hence 


f 


f=-^  =  A'R, 


A  being  the  angular  velocity. 

Hence  the  deviating  force  of  the  mass  m 


=  mf=  fft  p  =  fnA*r, 


and  is  equal  and  opposite  to  the  centrifugal  force. 

Again,  if  a  solid  body  of  mass  M  revolve  with  an  angular 
velocity  A  about  an  axis  passing  through  its  C.  of  G.,  the  total 


'.V- 


CENTRIFUGAL  FORCE. 


183 


centrifugal  force  will  be  nil,  provided  the  axis  of  rotation  is  an 
axis  of  symmetry,  or  is  one  of  the  principal  axes  of  inertia  at 
the  C.  of  G. 

If  the  axis  of  rotation  is  parallel  to  one  of  these  axes,  but 

at  a  distance  R  from  the  C.  of  G., 


the  centrifu- 
gal force 


W    ..^ 


\  =  ^mrA*  =  A'2mr  =  A'MR  =  —  A'R, 


r  being  the  distance  of  a  particle  of  mass  m  from  the  axis,  and 
W^the  weight  of  the  body.  Thus  the  centrifugal  force  is  the 
same  as  if  the  whole  mass  were  concentrated  at  the  C.  of  G. 

If  the  axis  of  rotation  is  inclined  at  an  angle  6*  to  the  prin- 
cipal   axis,   the   body  will    be    con- 
stantly subjected    to   the  action   of 
a  couple   of   moment    2E  tan  6^,  E 
being  the  actual  energy  of  the  body. 

Example. — A  ring  of  radius  r 
rotates  with  angular  velocity  A  about 
its  centre  0.  Let  p  be  the  weight  of 
the  ring  per  unit  of  length  of  periph- 
ery. Consider  any  half-ring  AFB. 
The  centrifugal  force  of  any  element 

CC^^-^A^r. 
g 
The  component  of  this  force  parallel  to  AB,  is  balanced  by 

an  equal  and  opposite  force  at  C" ,  the  angle  COB  being  = 

the  angle  CO  A.     Thus  the  total  centrifugal  force  parallel  to 

AOB\%  nil. 

The  component  of  the  force  at  (7,  perpendicular  to  AB, 


./.^A\  sin  COD=^^—-A'r  cos  C'CE 
g  g 


Fig.  19s. 


g         CC     '^  g 


W 


-.!■! 


\\\ 


?4| 


rfl 


i 


^  1: 

I 


'■iil 


I 


I 


I 


\'  u 


1411 


I  '1 


I  !*  ■■   : 


184  THEORY  OF  STRUCTURES.  ' 

Hence,  the  total  centrifugal  force  perpendicular  to  AB 

If  T  is  the  force  developed  in  the  material  at  each  of  the 
points  A  and  B^ 


2  7'=  2^A'r\ 
g 


since  the  direction  of  T  is  evidently-  perpendicular  to  AB. 


.-.  T  =  -A'r'  =  ^v\ 
g  g 


V  being  the  circumferential  velocity. 

Let  /  be  the  intensity  of  stress  at  A  and  B,  and  w  the 
specific  weight  of  the  material. 

Assuming  that  T  is  distributed  uniformly  over  the  sectional 
areas  at  A  and  B, 


XV 

g 


if-,  K 


Thus,  the  stress  is  independent  of  the  radius  for  a  given 
value  of  V,  and  the  result  is  applicable  to  every  point  of  a  flex- 
ible element,  whatever  may  be  the  form  of  the  surfaces  over 
which  it  is  stretched. 

22.  Impact. — When  a  body  strikes  a  structure,  or  member 
of  a  structure,  the  energy  of  the  blow  is  expended  in 

(i)  overcoming  the  resistance  to  motion  of  the  body  struck ; 

(2)  deforming  the  body  struck  ; 

(3)  the  kinetic  energy  of  either  or  of  both  of  the  bodies 
after  impact,  if  the  motion  is  sensible  ; 

(4)  deforming  the  striking  body; 

(5)  producing  vibrations. 


IMPACT. 


185 


Generally  speaking,  the  energy  represented  by  (5)  is  very 
small  and  may  be  disregarded.  Also,  if  the  striking  body  is 
very  hard,  the  energy  (4),  absorbed  in  its  deformation,  is  inap- 
preciable and  may  be  neglected. 

First,  let  a  body  of  weight  P  fall  through  a  vertical  dis- 
tance h  and  strike  a  second  body,  the  point  of  application 
moving  in  the  direction  of  the  blow  through  a  distance  x 
against  a  mean  resistance  R'.    Then 

Piji  -\-  x)=z  work  done  =  R'x. 

Let  Fbe  the  velocity  of  the  striking  body  at  the  moment 
of  impact.    Then 


P  V 

energy  of  blow  = =  Rx  =  P{Ji  -f-  x). 

S   2 


The  actual  resistance  is  directly  proportional  to  the  dis- 
tance through  which  the  point  of  application  moves,  so  long  as 
the  limit  of  elasticity  is  not  exceeded.     Its  initial  value  is  nil, 

and  if  R  is  its  max.  value,  the  mean  value  '\s  R'  =  — . 


r 

1  i 


f  1 


NHi 


]  !  >  )»  ^ 

•p;vi. 


t'li  *!h 


•'    I 


», .  |.| 


*    .8 
i    5 


I ; 


f    1 


,if 


PV'      Rx 
.-.  -7-=  Y=^(^^+^)- 


\i  h  =  o,  R  =  2P,  or  the  sudden  application  of  a  load  P 
from  rest,  produces  a  pressure  equal  to  twice  the  load,  pro- 
vided the  limit  of  elasticity  is  not  exceeded. 

Example.  A  i-oz.  bullet  moving  with  a  velocity  of  800  ft. 
per  sec.  strikes  a  target  and  is  stopped  dead  in  the  space  of 
-^(j  inch  (^  =  32).    Then 


i.-iV.^.(8oo)'  =  /?'./^.-iV; 


>     II 


"itmBMM-l 


:?r:.:  ^i 


1 86 


THEORY  OF  STRUCTURES. 


.'.  R',  the  mean  resistance  overcome  by  the  bullet,  =  5000  lbs. 
The  time  in  which  the  bullet  is  brought  to  rest 


momentum      iV  •  tjV  •  800 


force 


5000 


3200 


sec. 


Ni'xf,  let  a  body  of  weight  IV,  moving  in  a  given  direction 
with  a  velocity  7\  strike  a  body  of  weight  W,  moving  in  the 
same  direction  with  a  velocity  ?', .  After  impact  let  the  bodies 
continue  to  move  in  the  same  direction  with  a  common  ve- 
locity V, 

w        vv 

— ^i\  4"  — ^^'a  =  momentum  before  impact 


momentum  after  impact 

W      Wy 


or 


rr.r,  +  w,v,  =  {w,+  w,)v. 


Energy  before  impact  =  — '  -'-H -'  —  . 

A'"     2  I'"     2 


after 


=  ( 


yy.  +  n\  w 

g  /2 


Energy  lost  by  impact 


-(^F..-'+^F,TO--(^F,  +  rr,) 


If  ei 

must  b( 
energy 

EXA 

liead  of 

l;  round 

pile  beii 

Let  : 

"    / 

"    ^ 
«   V 


Px 


Also,  ( 
and  then 


Again, 


Finally 


m 


IMPACT. 


187 


If  either  of  the  bodies  is  subjected  to  any  constraint,  energy 
iTuist  be  expended  to  overcome  such  constraint,  and  the  loss  of 
energy  by  impact  will  be  less. 

EXAMI'LIC  I.  Let  a  weight  of  W^  tons  fall  h  ft.  upon  the 
head  of  a  pile  weighing  W^  tons  and  drive  it  a  ft.  into  the 
i;round  against  a  mean  resistance  of  R  tons,  the  head  of  the 
pile  being  crushed  for  an  appreciable  length  x  ft. 

Let  V  be  the  velocity  of  the  weight  when  it  strikes  the  pile  ; 
mean  force  of  the  blow  ; 

distance  through  which  pile  moves  during  ac- 
tion of  blow ; 
duration  of  the  blow  in  seconds ; 
common  velocity  of  the  pile  and  weight  during 
action  of  blow ; 
'  XT    "     "     distance  through  which  pile  moves  after  the 

blow. 
Px  -f-  Ry  =  work  done    in  crushing   the  pile  -f-  work 
done  in   overcoming  ground-resistance 
in  time  t  =  energy  dissipated  by  blow 


y     <•       «' 
t     "       " 


=  W./i 


IV,  +  w,  r 


(I) 


Also,  considering  the  change  of  momentum  first  of  weight 
and  then  of  pile, 


W  W 

Pt  =  — '(z;  -  F)  =  y?^  4-  — '  V. 

s  s 


(2) 


Again, 


Rz  =  work  done  after  blow  =  — ^~ — ?  — .      .    (2) 

g         2  ^^' 


l<ll 


Finally,  J'  +  «  =  «, (4) 

and 

v'-=2gh (5) 


l88  '/'//EOA'V  OF  STKUCTUKKS. 

Thus,  if  W^^  rr, ,  //,  a,  and  x  arc  known,  cqs.  (i)  to  (5)  will 
give  P,  t,  R,  y,  r,  and  s. 

Ex.  2.  Let  a  hammer  weighing  W^  lbs.  moving  with  a  ve- 
locity of  x<  ft.  per  sec.,  strike  a  nail  weighing  W^  lbs.  and  drive 
it  X  ft.  into  a  piece  of  timber,  of  weight  W,,  against  a  mean 
resistance  of  A'  lbs. 

First,  assume  the  timber  to  be  fixed  in  position. 

Let  F,  be  the  common  velocity  acquired  by  the  hammer 
and  nail. 


F" 
(JF,  4-  If,)-  -  =  energy  expended  in  overcoming  R 


But 


W  W  -!    W 

—'V  —  change  of  momentum  =  — —'- — -'F, .      .    (2) 

o  o 


W^  V" 


W,  +  IF,  2i^ 


Rx, 


(3) 


W,v 


and  the  time  of  the  penetration  =  77-.-  --  —-  sec.      ...     (4) 

Seeond,  let  the  timber  be  free  to  move,  and  let  F,  be  the 
common  velocity  acquired  by  the  hammer,  nail,  and  timber. 


(W^.+  f^)^ 


But 


energy  expended  in  overcoming  R  plus 
the  energy  expended  in  producing  the 
velocity  F, 

Rx  +  {W,+  ]V,+  W,)^.   ...     (5) 


w,       rF.+  rr        if.  +  iK  +  iK  „  ... 

— V  — F,  = F, .     .     .     (6) 


(;,V    /•///:    I.XT/iXSION   OF  A    PRISMATIC  liAK.  1 89 

llciicc,  substituting  these  values  of  F,  and  F,  in  cq.  (5), 


w:w. 


v 


also,  the  time  of  the  penetration 


--^Rx\    .     .     .    (7) 


iV,~  W,-\-W,+  W,gR 


.ill 

I:     'if! 


n 

J      tf,  ^ 


«fS 


and  the  distance  through  which  the  timber  moves 


V, 


w:  w. 


2  '  -  ( H\  +  W,  4-  H/ )'  ai'-A' 


•-7^:rr.,ft. 


(9) 


23.  On  the  Extension  of  a  Prismatic  Bar. — The  ele- 
mentary law  of  extension  is  sometimes  enunciated  as  follows: 

A  prismatic  bar  of  length  L  and  sectional  area  A  is 
stretched,  and  its  length  is  L  -\- x  when  the  force  of  extension 
is/';  if  ^//Ms  the  increment  of  force  corresponding  to  an  in- 
crement dx  of  length, 


1, 


dP=EA 


dx 


Hence,  the  force  producing  an  extension  /  is  equal  to 


X  ^"^i-TT^  =  ^^  ^°S.(i  +  2/      ^"  '"PP^^"^- 


But 


^°g^('  +  z)  =  z  -  2-  (z)  -^-ilz)  -  •  •  •  =  Z '  ^PP''^^- 


.'.P,  =  EAj. 


Mti 


m^' 


I     t,. 


■  ! 

if!* 


.ii-...v-t 


i  ' 


V 


li 


190  THEORY  OF  STRUCTURES. 

^      ,r  ^  ,      ,  .        dP,       EA         ^  EA 

Corollary. — From  the  last  equation,    -7^  =    y-,  and  — 

is  consequently  a  measure  of  the  longitudinal  stiffness  of  a  bar, 
so  that  for  the  same  material,  the  stiffness  varies  directly  as  the 
sectional  area  and  inversely  as  the  length,  while  for  different 
materials  it  also  varies  directly  as  the  coefficient  of  elasticity. 
Work  of  Extension. — The  force  producing  the  increment  dx 

lias    for    its    least   value   P{=  EAj),  for  its  greatest  value 

dP 
P-\-dP,  and  for  its  mean  value  P-\ ,  so  that  the  work  done 


is 


(P-\-—-\dx  =  Pdx,  approximately. 


Hence  the  work  done  in  stretching  the  bar  until  its  length 
is  Z  -}-  /  is  equal  to 


/ 


Pdx 


pi       X  EA  P 

J^  EA%  dx  =  ^ 


L 


L  2' 


24.  On  the  Oscillatory  Motion  of  a  Weight  at  the  End 
r,  of  a  Vertical  Elastic  Rod. — An  elastic  rod  of  natu- 
ral length  L{OA)  and  sectional  area  A  is  suspended 
from  O,  and  carries  a  weight  P  at  its  lower  end,  which 
elongates  the  rod  until  its  length  is  OB  =  L -\-  I. 

Assume  that   the  mass  of  the  rod   as   compared 
with  P  is  sufficiently  small  to  be  disregarded,  then 


C, 
A 


/>=£/(£. 


If  the  weight  is  made  to  descend  to  a  point  C,  and 
is  then  left  free  to  return  to  its  state  of  equilibrium,  it 
must  necessarily  describe  a  series  of  vertical  oscilla- 
tions about  B  as  centre. 

Take  B  as  the  origin,  and  at  any  time  t  let  the 
^veight  be  at  i^  distant  x  from  B;  also  let  BC  =  c. 
Two  cases  may  be  considered. 

First,  suppose  the  end  of  the  rod  to  be  gradually  forced 
duwu  to  C  and  then  suddenly  released. 


M 


Fig.  196. 


Acc( 


or 

and  henc 

f,t 

Now 
above  B 
Again 


and  integi 


and  the  os 
When. 


and  the  tir 


OSCILLATORY  MOTION.  I9I 

According  to  the  principle  of  the  conservation  of  energy, 

( — r )   =  the  work  done  between  C  and  M 

£  2    \dtl 


g 


_  EA  le     x\ 


or 


PildxV      Pi 

and  hence 

V,  the  velocity  of  the  weight  at  M,  =  a/S^Cc^  _  x^)i . 

Now  V  is  zero  when  x  =  ±  c,  so  that  the  weight  wili  rise 
above  B  to  a.  point  C^  where  BC^ ,  —  s  =z  BC. 
Again,  from  the  last  equation, 

^^\j  I  -{e-  x'y 

and  integrating  between  the  limits  o  and  x, 

/y/f  =  sin-^, 

and  the  oscillations  are  therefore  isochronous. 
When  x  =.  c, 

It      IT 
and  the  time  of  a  complete  oscillation  is 


\ 


i  ! 


'■  I,  i 
!i  .'^ 


Hf? 


I 


■\ 


m^i- 


ir 


192 


THEORY  OF  STRUCTURES. 


Next,  suppose  the  oscillatory  motion  to  be  caused  by  a 
weiffht  /'  falliii|:j  without  friction  from  a  point  D,  and  beini; 
suddenly  checked  and  held  by  a  catch  at  the  lower  end  of  the 
rod. 

Take  the  same  origin  and  data  as  before,  and  let  AD  =  h. 

The  elastic  resistance  of  the  rod  at  the  time  t  is 


EA 


so  tha 
oscillat 


Cor 


and  the  equation  of  motion  of  the  weight  is 


and  hen 


Pd\v 


l-\-x 


/r  de  L 


=  />-y(/+,r)=-P7; 


d\v  g 


Integrating, 


./.r\^ 


// 


j   =  —  "4.1-" -\-  c^,  c^  being  a  constant  of  integration. 


dx  g 

But  -J-  is  zero  when  x  =  c,  and  c^  =  rc'. 


Hence 


ri)'=f(--)  = 


This  is  precisely  the  same  equation  as  was  obtained  in  the 
first  case,  and  between  the  limits  o  and  x 


a/t  =  ''"'-' 7- 


or 


Cor.  : 

the    rod 

oscilJatioi 

Cor.  3 

Z  +  /,  wl 

to  its  elas 

motion. 

lias  reachi 

spending 

distance  a 

corresponi 

force  agai 

may  be  co 

however  si 

intervals,  t 

but  the  inc 

tiiiually. 

picx  on  ac( 

^fluctiiatio 

than  the  st 


mm 


1     VII 


OSCILLA  TOR  Y  A/0  TION. 


•93 


SO  that  the  motion  is  isochronous,  and  the  time  of  a  complete 
oscillation  is 


Cor.  I.  When  x  =  —  I. 


and  hence 


"vl- 


ri)"=-". 


I  (C»  -  /■')  =  2gh, 


or 


Cor.  2.  If  h  =  o,  i.e.,  if  the  weight  is  merely  placed  upon 
tlic  rod  at  the  end  A,  c  =  ±  /,  and  the  amplitude  of  the 
oscillation  is  twice  the  statical  elongation  due  to  /-*. 

Cor.  3.  The  rod  may  be  safely  stretched  until  its  length  is 
L-\-  /,  while  a  further  elongation  c  might  prove  most  injurious 
to  its  elasticity,  which  shows  the  detrimental  effect  of  vibratory 
motion.  If  a  small  downward  force  Q  is  applied  to  /'when  it 
has  reached  the  end  of  its  vibration,  it  will  produce  a  corre- 
sponding descent,  and  the  weight  P  will  then  ascend  an  equal 
distance  above  its  neutral  position.  At  the  end  of  the  interval 
corresponding  to  P's  natural  period  of  vibration,  apply  the 
force  again,  and  P  will  descend  still  further.  This  process 
may  be  continued  indefinitely,  until  at  last  rupture  takes  place, 
however  small  Pa.nd  (2  may  be.  If  Q  is  applied  at  irregular 
intervals,  the  amplitude  of  the  oscillations  will  still  be  increased, 
but  the  increase  will  be  followed  by  a  decrease,  and  so  on  con- 
tinually. In  practice  the  problem  becomes  much  more  com- 
plex on  account  of  local  conditions,  but  experience  shows  that 
7i  fluctuation  of  stress  is  always  more  injurious  to  a  structure 
than  the  stress  due  to  the  maximum  load,  and  that  the  injury 


194 


THEORY  OF  STRUCTURES. 


is  aggravated  as  the  periods  of  fluctuation  and  of  vibration  of 
the  structure  become  more  nearly  synchronous. 

An  example  of  a  fluctuating  load  is  a  procession  marching 
in  time  across  a  suspension-bridge,  which  may  strain  it  far 
more  severely  than  a  much  greater  dead  load,  and  may  set  up 
a  synchronous  vibration  which  may  prove  absolutely  dangerous. 
In  fact,  a  bridge  has  been  known  to  fail  from  this  cause. 

Cor,  4.  The  coefficient  of  elasticity  of  the  rod  may  be  ap- 
proximately found  by  means  of  the  formula 


"VJ 


T  being  the  time  of  a  complete  oscillation.     For  suppose  that 
the  rod  emits  a  musical  note  of  n  vibrations  per  second,  then 

V  ^  2« 

is  the  time  of  travel  from  C  to  C, ; 


m 


liv 


«  sS 


;  li 


.i   t 


,',  I  =  —5—5 ,     and  hence     E  = 


^it  n 


A      g 


Cor.  5.  Sujjpose  that  the  weight  is  perfectly  free  to  slide 
along  the  rod.  When  it  returns  to  A,  it  will  leave  the  end  of 
the  rod  and  rise  with  a  certain  initial  velocity.  This  velocity 
is  evidently  V2gh,  and  the  weight  accordingly  ascends  to  D, 
then  falls  again,  repeats  the  former  operation,  and  so  on.  The 
equations  of  motion  are  in  this  case  only  true  for  values  of  x 
between  x  = -\-  c  and  x  =^  —  I. 

25.  On  the  Oscillatory  Motion  of  a  Weight  at  the  End 
of  a  Vertical  Elastic  Rod  of  Appreciable  Mass. — Suppose 
the  mass  of  the  rod  to  be  taken  into  account,  and  assume : 

{a)  That  all  the  particles  of  the  rod  move  in  directions  par- 
allel to  the  axis  of  the  rod. 


Hence 


OSCILLATORY  MOTION. 


195 


~  ^°  +  /V^^^A'  +  /> 4-  p.gAx  =  o. 


(^)  That  all  the  particles,  which  at  any  instant  are  in  a  plane 
perpendicular  to  the  axis,  remain  in  that  plane  at  all  times. 

As  before,  the  rod  O/l  of  natural  length  L  and  sectional 
area  A  is  fixed  at  O  and  carries  a  weight  /\  at  A. 

Take  O  as  the  origin,  and  let  OX  be  the  axis  of  the  rod. 

Let  S,  S  -{-  </S,  and  x,  x  -\-  dx,  be  respectively  the  actual 
and  natural  distances  from  O  of  the  two  consecutive 
sections  MM,  M'M'.  q 

Let  p„  be  the  natural  density  of  the  rod,  and  p 
the  density  of  the  section  MM,  distant  B  from  O. 

The  forces  which  act  upon  the  rod   .  e  : 

{a)  The  upward  and  constant  force  .  \,dX  0. 

{!))  The  weight  P,  at  A. 

{c)  The  weight  of  the  rod. 

{(l)  A  force  X  per  unit  of  mass  through  the  slice 
hounded  by  the  planes  MM,  M'M',  distant  S  and 
S  -j-  '1^1  respectively,  from  O. 

Suppose  the  rod,  after  equilibrium  has  been  es- 
tablished, to  be  cut  at  the  plane  M'M'.     In  order  to 
maintain  the  equilibrium  of  the  portion  OM'M'  it 
will  be  necessary  to  apply  to  the  surface  of  this  plane  a  certain 
force  P,  and  the  equation  of  equilibrium  becomes 


M 
1/ 


Fig.  197. 


<  r 


u 


> 


1 '. 


\'-lh 


But  if  the  thickness  dS  of  the  slice  MM'  is  indefinitely 
diminished,  P  is  evidently  the  elastic  reaction,  and  its  value  is 


dB,—dx       t:  A  l^^         \ 

^^—dr-  =  ^'^\^x-'V 


Hence 


-  P„  +  ^pAXdH  +  EA  {~^_  -1)4-  p„gAx  =  9. 


!  1  ' 


TH 


'T"  n": 


196  THEORY  OF  STRUCTUREH. 

Differentiating  with  respect  to  x. 


dB,  d'S, 

pAX-r-  +  £A-r^  +  p,gA  =  o. 


dx 


dx" 


a, I' 
\\\ 


But  pdB  =  p^dx, 


or 


.'.  p,AX-\-  EA  j^  +  p,gA  =  o, 


Also,  p„AXdx  is  the  resistance  to  acceleration  arising  from 
the  inertia  of  the  slice,  and  is  therefore  equal  to 


-p^Adx-^, 


so  that 


X=- 


d^ 
'df  ' 


Hence 


d'B     Ecea 

df  ~  p„  dx" 


■\-g- 


(I) 


To  solve  this  equation. — In  the  state  of  equilibrium, 


-(J!-) 


is  the   tension   in   the   section   of   which   the   distance   from 


OSCILLATORY  MOTION. 


197 


0  is  X,  and  counterbalances  the  weight  P^  and   the  weight 
ft^A{l  —  x)g  of  the  portion  AMN  of  the  red. 

.-.  £^  (^  -  l)  =  />,  +  p^g(l  -  x), 


vf  f  i 


% 


or 


Integrating, 


S='+^+¥('--)- 


5  =  .+  A,+P^(,,_£).      ...     (3) 


There  is  no  constant  of  integration,  as  x  and  $  vanish 
together. 

This  value  of  ^  is  a  particular  solution  of  (i),  and  is  inde- 
pendent of  /. 


Put 


^  =  *  +  Ei-  +  ¥('--T)+-- 


■  W, 


I     ,^ 

wm 

'''  Mil 

• 

■  !  ■ "?' 

^^  being  a  new  function  of  x  and  /.    Then 


d^a 


dx' -     B^'^dx''  ^""^    de  ~  dt" 


Hence,  from  eq.  (i). 


d^z  _  E  (£z 
de  ~  p,dx^ 


d^s 


P.' 


The  integral  of  this  equation  is  of  the  form 


s  =  F{x^  v,t)  +/(.r  -  v,t), 


! 

jiiill 

198 


THEORY  OF  STRUCTURES. 


^^(=vf)^"'"^ 


ig  the  velocity  of  propagation  of  the  vibrations. 
The  full  solution  of  (i)  is  therefore  of  the  form 


26.  Inertia — Balancing. — Newton's  First  Law  of  Motioiv 
called  also  i\\c  Law  of  Inertia,  states  that  "a  body  will  continue 
in  a  state  of  rest  or  of  uniform  motion  in  a  straight  line  unless 
it  is  made  to  change  that  state  by  external  forces." 

This  property  of  resisting  a  change  of  state  is  termed 
inertia,  and  in  dynamics  is  always  employed  to  measure  the 
quantity  of  matter  contained  in  a  body,  i.e.,  its  mass,  to 
which  the  inertia  must  be  necessarily  proportional.  Thus,  to 
induce  motion  in  a  body,  energy  must  be  expended,  and  must 
again  be  absorbed  before  it  can  be  brought  to  rest.  The  inertia 
of  the  reciprocating  parts  of  a  machine  may  therefore  heavily 
strain  the  framework,  which  should  be  bolted  to  a  firm  foun- 
dation, or  must  be  sufificiently  massive  to  counteract  by  its 
weight  the  otherwise  unbalanced  forces. 

Example  i.  Consider  the  case  of  a  direct-acting  horizontal 

steam-engine.  Fig.  198.  At  any 
given  instant  let  the  crank  OP 
and  the  connecting-rod  CP  make 
angles  B  and  0,  respectively,  with 
the  line  of  stroke  AB. 

Let  V  be  ':he  velocity  of  the 
crank-pin  centre  P,  and  let  u  be 
the  corresponding  piston  velocity,  which  must  evidently  be  the 
same  as  that  of  the  end  C of  the  connecti.ig-rod. 

Let  CP  produced  meet  the  vertical  through  C  in  /. 
At  the  moment  under  consideration,  the  points  C  and  Pare 
turning  about  /as  an  instantaneous  centre. 


Fig.  198. 


U 
V 


IC 

IP 


sin  (^+0) 
cos  (J> 


INER  TIA  —BALA  NCING, 


199 


Let  W  be  the  weight  of  the  reciprocating  parts,  i.e.,  the 
piston-head,  piston-rod,  cross-head  (or  motion-block),  and  a  por- 
tion of  the  connecting-rod. 

Assume  (i)  that  the  motion  of  the  crank-pin  centre  is  uni- 
form; 
(2)  that  the  obHquity  of  the  connecting-rod  may- 
be  disregarded   without   sensible  error,  and 
.-,  0  =  0. 
Draw  PN  perpendicular  to  AB,  and  let  ON=x\  ON  is 
equal  to  the  distance   of  the  piston  from  the  centre  of  the 
stroke,  corresponding  to  the  position  OP  of  the  crank. 
The  kinetic  energy  of  the  reciprocating  parts 


:t 


id;. 


W^i^_W_ v"  sin'  e      W v' 
g   '^~  g        2 


r  being  the  radius  OP. 

.".  the  change  of  kinetic  energy,  or  work  done,  corresponding 
to  the  values  x^ ,  x^  of  x. 


■mm 


W  %!"  (X^  -  ^„»^ 
g 


r^^i- 


'^■lU 


Let  R  be  the  mean  pressure  which,  acting  during  the  same 
interval,  would  do  the  same  work.     Then 


W  z>'  x;  -  x," 
y  2  '     r' 


=  R{x,  —  x^), 


and 


,  Wv^x.  +  x, 

g   2        r' 


Hence,  in   the  limit,  when  the  interval   is  indefinitely  small, 
x^  =  .r,  =  X,  and  the  pressure  corresponding  to  x  becomes 


R  = -X. 

g  r 


,1 


. 


200 


THEORY  OF  STRUCTURES. 


This  is  the  pressure  due  to  inertia,  and  may  be  written  in  the 
form 


R=C-, 
r 


C  {— )  being  the  centrifugal  force  of  W  assumed  con- 

centrated  at  the  crank-pin  centre.  /?  is  a  maximum  and  equal 
to  C  when  x  =^  r,  i.e.,  ?.t  the  points  A,  B,  and  its  value  at 
intermediate  points  may  be  represented  by  the  vertical  ordi- 
nates  to  AB  from  the  straight  line  EOF  drawn  so  that 
A£  =  BF  =  C  In  low-speed  engines,  C  may  be  so  small  that 
the  effect  of  inertia  may  be  disregarded,  but  in  quick-running 
engines,  C  may  become  very  large  and  the  inertia  of  the  recip- 
rocating parts  may  give  rise  to  excessive  strains. 

Another  force  acting  upon  the  crank-shaft  is  the  centrifu- 
gal force  of  the  crank,  crank-pin,  and  of  that  portion  of  the  con- 
necting-rod which  may  be  supposed  to  rotate  with  the  crank- 
pin. 

Let  w  be  the  weight  of  the  mass  concentrated  at  the  crank- 
pin  centre  which  will  produce  the  same  centrifugal  force  as 
these  rotating  pieces  (i.e.,  wr  =  sum  of  products  of  the  weights 
of  the  several  pieces  into  the  distances  of  their  centres  of  gravity 
from  O). 

The  centrifugal  force  of  w  = . 

g   r 


Thus  the  total  maximum  pressure  on  the  crank-shaft 


WV^  V* 


=  C+~j:=~{lV-{-zv)  =  r{W  +  zv)j 


A  being  the  uniform  angular  velocity  of  the  crank-pin. 

This  pressure  may  be  counteracted  by  placing  a  suitable 
balance-weight  (or  weights)  in  such  a  position  as  to  develop  in 
the  opposite  direction  a  centrifugal  force  of  equal  magnitude. 


Let 


or 


from  whi 

Durir 

represent 

reciproca 

triangle  I 

when  the 

Durin 

reciprocal 

the  piston 

direction. 

InAE 

of  the  rec 

^?,  AE  rep 

Draw  E'C 

During 

E'O'  reprt 

celerate 

retarded 

The  ca 

N.B.— 

follows : 

u  =  V 


Wd_ 
g  d 


%: 


INER  TIA—BA  LA  NCING.  201 

Let  W^  be  such  a  weight  and  R  its  distance  from  0.    Then 


or 


RW,  =  r{W+w), 


m 

^^rfll 

H 

Hffli 

? 

,r  ■ 

''!;|j 

s 

i 

si 

t. 

1 

f 

■  r 

1 

from  which,  if  R  is  given,  W^  may  be  obtained. 

During  the  first  half  of  the  stroke  an  amount  of  energy 
represented  by  the  triangle  AEO  is  absorbed  in  accelerating  the 
reciprocating  parts,  and  the  same  amount,  represented  by  the 
triangle  BOF,  is  given  out  during  the  second  half  of  the  stroke 
when  the  reciprocating  parts  are  being  retarded. 

During  the  up-stroke  of  a  vertical  engine  the  weights  of  the 
reciprocating  parts  act  in  a  direction  opposite  to  the  motion  of 
the  piston,  while  during  the  down-stroke  they  act  in  the  same 
direction. 

\r\AE  produced  (Fig.  199)  take  EE'  to  represent  the  weight 
of  the  reciprocating  parts  on  the  same  scale  g- 
as  AE  represents  the  pressure  due  to  inertia.  E 
Draw  E'O'F'  parallel  to  EOF.  A 

During   the   up-stroke  the  orrlinates   of 
E'O'  represent  the  pressures  required  to  ac- 
celerate the  reciprocating  parts,  the  pressures  while  they  are 
retarded  being  represented  by  the  ordinates  of  O'F' . 

The  case  is  exactly  reversed  in  the  down-stroke. 

X 

N.B. — The   formula  R  =  C-  may  be  easily  deduced  as 

T 

follows : 

r.      ,  ,        .  ^«  ^d9      v" 

«  =  t;  sin  6',  the  acceleration  =  -y-  =1  v  cos  6  —-  =  -x  \ 

at  at      r 

W  du 
.:  —  '77  —  accelerating  force  =  force  due  to  inertia 

-!Z^-  - rf 

—   ^  r'^~      r' 


ml 


^''IH 


202 


THEORY  OF  STRUCTURES. 


Ex.  2.  Consider  a  double-cylinder  engine  with  two  cranks 
at  right  angles,  and  let  d  be  the  distance  between  the  centre 
lines  o^  the  cylinders  (Fig.  200). 


f  C.COS  Q 


.(%ntre  line 
of  Cylr. 


Y      Centre  line 


iC.Blne        o'Cylr. 


Fig.  200. 


The  pressures  due  to  inertia  transmitted  to  the  crank-pins 
when  one  of  the  cranks  makes  an  angle  B  with  the  line  of 
stroke  are 

P,—  C cos  0    and    P^=  C sin  B. 

These  are  equivalent  to  a  single  alternating  force 

P  =  C(cos  B  ±  sin  B) 

acting  half-way  between  the  lines  of  stroke,  together  with  a 
couple  of  moment 

M=P-=  C-(cos  B  ±  sin  B). 

2  2 

The  force  and  couple  are  twice  reversed  in  each  revolution, 
and  their  maximum  values  are 


Cd  ^ 

P.na..  =  CV2     and     M^^,  =  -J  ^• 


...-■]'    i    .i"o\d  the  evils  that  might  result  from  the  action 
of  the       '  (  couple  at  high  speeds,  suitable  weights  are 

introduced  in  such  positions  that  the  centrifugal  forces  due  to 


their  ro 

For  ex; 

upon  thi 

ing-whei 

Let  ; 

diametri 

crank-pii 

from  the 

Let  e 

the  balar 

Thee 


and  this  f 


Fig. 

forces  eacl 
the  angle  1 
that  betwe 
ant  couple 
the  axes  o 
to  the  line 
Q  and  < 

2F  CO'. 

and 

Fe  sin 


IliiJ 


INER  TIA—BA  LA  NCING. 


203 


their  rotation  tend  to  balance  both  the  force  and  the  couple. 
For  example,  the   weights  may  be  placed 
upon  the  fly-wheels,  or  again,  upon  the  driv- 
ing-wheels of  a  locomotive. 

Let  a  balance-weight  Q  be  placed  nearly 
diametrically  opposite  to  the  centre  of  each 
crank-pin  (Fig.  201),  and  let  R  be  the  distance 
from  the  axis  to  the  centre  of  gravity  of  Q. 

Let  e  be  the  horizontal  distance  between 
the  balance-weights. 

The  centrifugal  force  7^  due  to  the  rotation  of  Q 


Fig.  201. 


™ 

^Wm  H 

■ 

R 

\ 

K 

1    '      M' 

;mp, 

>\k 

1 
^  p 

, 

-til 

iVt  ' 

jiiiiiiiii 

'rrrfTr 

% 

if 

e  (velocity  of  Qf 
g 


R 


and  this  force  F  is  equivalent  to  a  single  force  F acting  half-way 
between  the  weights  and  to  a  couple  of  moment 


> 

\  90' A  90°; 

\ 

'\y 

Let  0  be  the  angle  between  the  radius 


Fig.  202. 


2 

to  a  balance-weight,  and  the  common  bisector 

of  the  angle  between  the  two  cranks  (Fig.  202). 

Since  there  are  two  weights  Q,  there  will 

g 
be  two  couples  each  of  moment  F  - ,  and  two 


forces  each  equal  to  F  acting  half-way  between  the  weights, 
the  angle  between  the  axes  of  the  couples  being  1 80°—  20,  and 
that  between  the  forces  being  20.  The  moment  of  the  result- 
ant couple  is  Fe  sin  0,  and  its  axis  bisects  the  angle  between 
the  axes  of  the  separate  couples ;  the  resultant  force  parallel 
to  the  line  of  stroke  =  2/^  cos  0. 

Q  and  0  may  now  be  chosen  so  that 

2F  cos  0  =  maximum  alternating  force  =  C  V2, 
and 

Fe  sin  0  =  maximum  alternating  couple  =  —  V2. 

d 
.'.  tan  0  =  -  , 


111 


■f :4r   ' 


i;:: 


:!  ' 


1  .  ri 

hi 


I 


304 
and 

or 

and 


THEORY  OF  STRUCTURES. 


gr"  g  r  ey 


e'-\-d* 


^'  +  ^' 


Ex.  3.  Again,  the   pressure   d7  at  a  dead  point  may  be 
balanced  by  a  weight  Q  diametrically  opposite. 
If  R  is  the  radius  of  the  weight-circle,  then 


g  r  g  r" 


and 


,'.Q^W 


R  ' 


e-i-d 
The  weight  Q  may  be  replaced  by  a  weight  Q—~—  on  the 


2e 


e  —  d 
near  and  a  weight  Q—z —  on  the  far  wheel.     Thus,  since  the 

cranks  are  at  right  angles,  there  will  be  two  weights  90°  apart 

e  -^  d  e d 

on  each  wheel,  viz.,  Q in  line  with  the  crank  and  Q . 

These  two  weights,  again,  may  be  replaced  by  a  single  weight 
B  whose  centrifugal  force  is  the  resultant  of  the  centrifugal 
forces  of  the  two  weights.    Thus 

IB 7ry_  (Q  ej±_d vy    iq e^-d v_y 

^g   R^~^g     2e      Rl'^^g      2e     Rl' 

v'  being  the  linear  velocity  at  the  circumference  of  the  weight- 
circle. 


.-.  ^  =  (2' 


2^' 


or 


If 


a  1 


Note.- 
nil,  and  B 

27.  Ct 

Ex.  I. 


Let  CP 
in  OP  take 

The  pis 
centre  are  c 


If  the  veloc 
by  OP,  then 
velocity  u. 


CURVES  OF  PISTON   VELOCITY. 


205 


or 


B 


Q    A'  4-  cV 


If  a  is  the  angle  between  the  radius  to  the  greater  weight 

e  +  d 

Q and  the  crank  radius, 

2e 

Q  e-d  v^l 

g     2c      R       e  —  d 

Qej±_dv'^~e^d' 

g     2e     R 

Note. — In  outside-cylinder  engines  e  —  dh  approximately 

nil,  and  ^  =  (2  =  W^. 

27.  Curves  of  Piston  Velocity.— Consider  the  engine  in 
Ex.  I. 

s 


Fig.  203. 


Let  CP  produced  intersect  the  vertical  through  O  in  T',  and 
in  CPtake  OT'  =  OT. 

The  piston  velocity  u  and  the  velocity  v  of  the  crank-pin 
centre  are  connected  by  the  relation 


u 

V 


sin  {0  H-  0)  _0T _  OT^ 
"OF 


cos  0 


OP 


.  .  .  (I) 


If  the  velocity  v  is  assumed  constant,  and  if  it  is  represented 
by  OP,  then  on  the  same  scale  OT'  will  represent  the  piston 
velocity  «.    Drawing  similar  lines  to  represent  the  value  of  u 


I 


i  ^K   I 


f 


:i 


1 1 


ill 


a 


\  \\\ 


1  'i 


') '  If 
If 


.  ! 


;i 


206 


THEORY  OF  STKUCTUKES. 


for  every  position  of  tlic  crank,  the  locus  of  'f  will  be  lountl  to 
consist  of  two  closed  curves  OGS,  OUT,  caWcd  X\\iz  polar  curies 
of  piston  velocity.  They  pass  through  the  point  O  and  through 
the  ends  5  and  T  of  the  vertical  diameter.  On  the  side  towards 
the  cylinder  they  lie  outside  the  circles  having  OS  and  OT  as 
diameters,  while  on  the  side  away  from  the  cylinder  they  lie 
inside  the  circles.  If  the  connecting-rod  is  so  long  that  its 
-obliquity  may  be  disregarded, 

0  =  0    and     u  =  t>  sin  6, 

and  the  curves  coincide  with  the  circles. 

A  rectangular  dxdigrdim  of  velocity  may  be  drawn  as  follows : 


C     M 


Upon  the  vertical  through  C,  Fig.  204,  take  CL  =  OT;  the 
locus  of  L  is  the  curve  required  for  one  stroke.  A  similar 
curve  may  be  drawn  for  the  return  stroke  either  below  A/N  ov 
upon  the  prolongation  A^R  {=  MN)  of  MN. 

If  the  obliquity  of  th«^  connecting-rod  is  neglected,  the 
curves  evidently  coincide  with  the  semicircles  upon  AIN  and 
AT?,  MN  (r=  NR)  defining  the  extreme  positions  of  C.  Tiic 
obliquity,  however,  causes  the  actual  curve  to  fall  above  the 
semicircle  during  the  first  half  of  the  stroke,  and  below  during 
the  second  half. 

Again,  let  the  connecting-rod  (/)  =  n  cranks  {f).     Then 


n. 


f  '     a   X          au       ^\          /  •     /J  I     sin  '9  cos  6*  \      ,  ^ 
//  =  V  (sm  p  -X-  cos  V  tan  0)  =  7'  (  sin  v  A —  ).    (2) 

\  Vn'  -  sin"  ej 


sin  ^       / 

sin  0  ""  r 

and  by  eq.  I, 

If  th( 
tan 


and 


28.  C 

j)ositii)ii  ( 


If  the 

resented  t 

resent   tht 

already  dr 

to  reprcsei 

If  tlie  pre 

usually  the 

radius,  rej. 

of  P.     Aft 

take  OP'  ii 

of  the  crai 

lesponding 

directly  oh 

parallel  to 

sent  the  rec 

may  be  dra 

29.  Cur 

curve  of  en 

position  01 

being  equal 


n   { 


CURVE   OF  CRANK-EFFORT— CURVES  OF  ENERGY.      20? 
If  the  obliquity  is  very  small, 


tan  0  =  sin  0  = 


SI 


in  e 


n 


,  approximately, 


and 


(  ■     ft   ^   sin^cos  ft]  i  .     n   ,    sin  2^ 


28.  Curve  of  Crank-effort. — The  crank-cjfort  F  for  any 
position  OP  o{  the  crank  is  tlie  component  along  the  tangent 
at  P  oi  the  thrust  along  the  connecting-rod. 


This  thrust  = 


cos  0 


^_^sin(^y  +  0) 
cos  0 


If  the  pressure /*  upon  the  piston  is  constant,  and  if  it  is  rep- 
resented by  OP,  then,  on  the  same  scale,  OT',  Fig.  203,  will  rep- 
resent the  crank-effort.  Thus,  the  curves  of  piston  velocity 
already  drawn  may  also  be  taken 
to  represent  curves  of  crank-effort. 
If  the  pressure  P  is  variable,  as  is 
usually  the  case,  let  OP,  the  crank 
radius,  represent  the  initial  value 
of  P.  After  expansion  has  begun, 
take  OP'  in  OP,  for  any  position  OP 
of  the  crank,  to  represent  the  cor- 
responding pressure  which  may  be 
ilirectly  obtained  from  the  indicator-diagram.  Draw  P'T' 
parallel  to  PT,  and  take  OT"  ^  OT.  Then  OT"  will  repre- 
sent the  required  crank-ef?ort,and  the  linear  and  polar  diagrams 
may  be  drawn  as  already  described. 

29.  Curves  of  Energy— Fluctuation  of  Energy. — In  the 
curve  of  crank-efYort  as  usually  drawn,  the  crank-effort  for  any 
position  OP  of  the  crank  is  the  ordinate  S'H,  the  abscissa  DH 
being  equal  to  the  arc  AP,  i.e.,  to  the  distance  traversed  by  the 


F"IG.  205. 


Hi 


m 


H 


:    :  ,iM 


^^B.') 

■  t ' 

■: 

■ 

1 

1'      ' 

t 

■ 

" 

:4- 
1  ■- 

-i, 

■ 

^^H 

I  ■ 

u[ 

■  f  ^-  ■ 

'< . 

V  v»;it? 

Mm 

(■  . 

tft 

m 

i 

208 


THEORY   OF  STRUCTURES. 


ii 


point  of  application  of  the  crank-effort.     Thus,  DSE  andEVG 
being  tlic  curves, 

DE  =  EG  =  semi-circumference  of  crank-circle  =  nr. 
If  the  obliquity  is  neglected,  the  curves  of  crank-effort  arc 
the  two  curves  of  sines  shown  by  the  dotted  lines. 


The  area  DS H  also  evidently  represents  the  zvork  done  as 
the  crank  moves  from  OA  to  OP,  and  the  total  work  done  is 
represented  by  the  area  DSE  in  the  forward  and  by  E  VG  in 
the  return  stroke. 

Let  F^  be  the  mean  crank-effort.     Then 

F,  X  2nr—  2PX  2r, 


assuming  Pto  be  constant. 


/^,= 


2P 

7t  ' 


2P 


Draw  the  horizontal  line  1234567  at  the  distance  —  from 

DEG,  and  intersecting  the  verticals  through  D,  E,  and  ^7  in  i, 
4,  and  7,  and  the  OJrves  in  2,  3,  5,  and  6.  The  engine  ma  be 
supposed  to  work  against  a  constant  resistance  R  equal  and 
opposite  to  the  mean  crank-effort  F„ . 

From  Z>  to  2,  7?  >  crank-effort,  and  the  speed  must  there- 
fore continually  diminish. 

From  2  to  3,  i?  <  crank-effort,  and  the  speed  must  contin- 
ually increase. 

Thus  2  is  a  point  of  min.  velocity,  and  therefore  also  of 
min.  kinetic  energy. 

From  3  to  £',  ^  >  crank-effort,  and  the  speed  must  contin- 
ually diminish. 


are  necessari 


"W^ 


CUFVES  OF  ENERGY— FLUCTUATION  OF  ENERGY.     209 

Thus  3  is  a  point  of  max.  velocity,  and  therefore  also  of 
max.  kinetic  energy. 

Similarly,  in  the  return  stroke,  5  and  6  are  pnnts  of  min. 
and  max.  velocity,  respectively. 

The  change  or  fluctuation  of  kinetic  energy  from  2  to  3  = 
area  283,  bounded  by  the  curve  and  by  23. 

The  fluctuation  from  3  to  5  =  area  3^5,  bounded  by  35  and 

by  the  curve. 

F      Fr 
Again,  since  -j^=l—  ^  the  ordinatcs  of  the  curves  may  be 

taken  to  represent  the  moments  of  crank-effort,  and  the  abscissae 
arc  then  the  corresponding  values  of  ^. 

The  work  done  between  A  and  any  other  position  P  of 
the  crank-pin 


db 


Jo  Jo  \  Vn^-  sin'  0) 

=  Pr{i  —  cos  ft  -\-  n  —  V'«"  -  sin""^). 

If  there  are  two  or  more  cranks,  the  ordinatcs  of  the  crank- 
effort  curve  will  be  equal  to  the  algebraic  sums  of  the  several 
crank-efTorts.  For  example,  if  the  two  cranks  are  at  right 
angles,  and  if  F, ,  F^  are  the  crank-efforts  when  one  of  the 
cranks  (/^,)  makes  an  angle  0  with  the  line  of  stroke, 


^,  =  4i„.  +  ?i'if^) 


and 


=H 


cos  6 


/?__}_/?=  P(sin  ^  +  cos  6^ 


sin  2^ 
2n    /■ 

combined  crank-effort. 


P  being  supposed  constant. 

Note. — In  the  case  of  the  polar  curves  of  crank-effort,  if  a 
circle  is  described  with  O  as  centre  and  a  radius  =  mean  crank- 

2P 

effort  =  — ,  it  will  intersect  the  curves  in  four  points,  which 

are  necessarily  points  of  max.  and  min.  velocity. 


m 


(■ ' 


:i 


m 


210 


THEORY  OF  STRUCTURES. 


w 

w 

H 

a 
< 

?: 
O 

OS 

o 

H 

O 

Q 

< 


u 

H 

< 


t/1 
H 

o 
z 

W 
Oi, 
H 


J3       O  u     • 

is   u'^.s 


0  rf 


O  O        C 
>S5       - 


a 

fl.  «  O  u  O  k. 


"=  il  5 

U  K  O 


8     8    8  8  8  8 

i/>      »o     m  in  lo  W1 

■8    "8  'S'S'8^ 


888i 

to  u>  in  c 


1  ^  m  en  in  in  H  ^ 


00    00  00  00  00  m -r  ■*  o 


a  c  c 
o  o  o 

N  rn  S» 


aacucQac 

oooooooo 


m^t 


\%. 


^1 


K 


c 


t/1 


^ 


I 
■a 


s  s 


m  in  (>co  oo  o6 


^  ^ 


«1 


8^ 


SiifeS 


*  . 

2  0 


Jo 
.•a 


uJS  »l  o 


it 
'5. 
u  Be 

5°  si 

(«  "Q  - 


o 

T3 


O 

•a 


a 

rt 


>    c  u 
c  M 

JO  tic 
bcu 
c  ^ 
rt  j^ 

_  w 

^  2 
a* 


3 
bg 
c 

u 

•a 
c 
rt 

□ 

9 
o 

u 

e 

rt 


rt  Wrt 
•S  c-a 


s 


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21 
a 

AM 


a 


u 


1^  IF  3 


01 

1 

(A 


Acacia. . 
■Mder.  ... 

Apple 

Ash,  Canadi 
Asli,  Enelis 
Heech..... 

Mirch 

Hux 

Hluegiiui. 

Cedar 

Clicstnut. . 

Kbony 

.lil.T:,  Canadi 


TABLES. 


THE  STRENGTHS,    ELASTICITIES,    AND    WEIGHTS   OF 
VARIOUS   ALLOYS.    ETC. 


SII 


Mutcriul. 


Aluminum 

Urass  

"     (li.immcred). 

Urass  wire    

Coppt-r-platc,  liammercd,. . . 

"  annealed 

(Upper  wire 

(iiiiimelal 

Lead     

Lead  wire 

rliosplior-bronze 

I'm    

Zinc 

Leather 


Max.  Load  on  OriKinal  Area 

in  lbs.  per  aq.  in. 

Tension. 

Com- 

Shcar- 

pression. 

"iK. 

aS.Soo 

17,600 

10,380 

53,000 

3o,o<x> 

58,000 

fKl,<KXl 

3(>,CX-HJ 

1,850 

7,100 

.i.>'» 

57.000 

4,y8o 

7.500 

4,000 

Young's 
Modulus, 

E 
(in  lbs.). 


9,600,000 

0,IfX>,000 

I  .;,ooo,oo<j 
14,oo.),oo(j 

I5,(KX),CXJ») 
l5,(K.(l,OCJO 
I7,(KXJ.(IOO 

9,yot>,ofX3 

7ro,(x>o 

9.^0, 000 

14,000,000 

5,69o,<xxi 

13,500,000 

25,000 


Coef- 
ficient of 
KlKidity, 
G        I 
(in  lbs.). 


^ciRlit 

in  lbs. 

per  cu  ft 


3,600,000     1 60  to  166 

3, 400, (XXI    487  to  534.4 
5,6ocj,ooo 


5,700,000 
5,700,000 


556 

555 
579 
71a 


5,750,(xxj 

t,i4o,o<xj     45'.  to  46S 
'  ^",   to  449 


THE  STRENGTHS,  ELASTICITIES,  AND  WEIGHTS  OF  TIMBERS. 

This  tabic  contains  the  results  of  the  most  recent  and 
most  reliable  experiments,  bui,  generally  speaking,  only  small 
specimens  of  the  material  have  been  testeci.  It  is  found  that 
the  strength,  elasticity,  and  weight  of  a  timber  are  affected  by 
the  soil,  age,  seasoning,  per  cent  of  moisture,  position  in  the 
log,  etc.,  and  hence  it  is  not  surprising  that  specimens  even 
Avhen  cut  out  of  the  same  log  show  results  which  often  differ 
very  widely  from  the  mean.  Additional  experiments  on  large 
timbers  are  needed,  and  in  each  case  should  be  accompanied 
b)'  a  complete  history  of  the  specimen  from  the  time  of  felling. 


Description  of 
Timber. 

Tensile 
Strength 

in  tons 
per  sq.  in. 

Com- 
pressive 
Strenifth 
in  tons 
per  sq.  in. 
alonu 
Fibres. 

Shearing 
Strenuth 

in  tons 
per  S(i.  in. 

alont; 

Fibres. 

Voi'nff'j 
Modulus, 

F. 
(in  tons). 

Coef- 
ficient 

of 

Ri«id- 

Coefficient 
of  Bcndin;; 

Strength 
in  tons 

per'sq.  in. 

Weiifht 
in  lbs. 

per 
cu.  ft. 

4.5  to  6.3 

8.8 

2.45 

5.35107.58 

4.9  to  9.8 

6.69 

9 

a  7 

a.  33  104.9 

4-3 

71 
3« 

25 

3fi7 
1.47  to  3.37 
4.6 
3.078 
a. ^6 

8.48 

3.9 

.2  to  .31a 
.35  to  .364 

.308 

6ao 
733 
607 
734 
803 
509 
ai7 

509 
1 100 

5.86 

Alder 

50 

so 

47 
43  to  53 
43  to  S' 
45  f'  *fi 

64 

35  to  47 
35  to  4« 

47 

Apple 

Ash,  Canadian.. 
Ash,  English  . . . 
Heech 

Hirch 

Hox 

Blue  (funi 

Cedar 

C:hestnut 

Kbony 

Kiai,  Canadian.. 

212 


THEORY  OF  STRUCTURES. 


THE    STRENGTHS,    ELASTICITIES,    AND   WEIGHTS    OF  TIMBERS 

{Continued.) 


Description  of 
Timber. 


Elm,  Eng 

Gieenlieart 

Hawthorn .. 

Hazel 

Hornbeam     . .  .. 

Iron  bark  

Ironwood 

Jarrah 

Lancewood 

Larch 

Lignum  vitJe. . .. 

Locust  

Mahogany, Span.. 
Hond 

Maple 

Mora 

♦Oak,  Am 

Oak,  Am.  red. . . 

"    white 

"    Eng 

Pine,  Dantzic... 

"     Memel 

"     pitch 

♦  "     red 

"     red 

♦  "     yellow  .. .. 
"     yellow.  . .. 

♦  "     white 

white  . .   . . 


Plane . 
Poplar 

♦  Spruce 

*  " 


Sycamore. 

Teak 

Walnut... 
Willow... 


Tensile 
Strength 

in  tons 
per  sq.  in. 


5-89 
2.7  to  4.1 
4.68 
8.48 
9.1 
7.12 
4-31 
■•3« 

3 . 6  to  6 . 7 
3.92  to  4.55 

5.26 

4.5  to6.7 

1.7  to  7.3 
1.3  103.6 
4-7107.7 

41 

4.46 

8.8 

5'4 

3-5 

4-5 

4.6 

1.7  to  6.67 

2.2  to  6.87 

1.3  to  5.1 
5-4 
2.94 

2.99  to  5.y7 
3 
5-8 
4.7  to  6.7 

3-5 

4.6  to  6.25 


Com- 
pressive 
Strength 
in  tons 
per  sq.  in. 
along 
Fibres. 


4.6 
4.46106.5 


2.6 
4-54 

5-21 

3-2 

1.42  to  2.45 
4.46 
1-33 
33 

2.23 
4.4 

1.89  to  2.6 
2.84 
4.4 


3'5 
2.4  to  3 
2.4  to  3.6 
2.24 


3.16 

5.35 

2.7 

••5 


Shearing 

Strength 

in  tons 

per  sq.  in. 

along 

Fibres. 


•  535 
33 


,324  to  .446 
335  'O  -43' 


.227 
119  to  .164 

,113  to  .167 


Young's 
Modulus, 

E 
(in  tons) 


759 


Coef- 
ficient 

of 
Rigid- 
ity, 
G. 


446 


830 
577 

664 
1025 

670 
962 

179 
900 
484 

604 
340 
594 
438  to  737 
700 
464 
1000 

629 


Coefficient 
of  Bending 

Strength 
in  tons 

per  sq.  in. 


S.15 


413 


560  to  1339 
712  to  879 


2  712 

4-55 


93 
7> 

255 

5' 

146 

03 


1.63  to  2.86 


Weight 
in  lbs. 

per 
cu.  ft. 


34  to  37 
58  to  72 


47J 


42  to  63 
32  to  38 
41  to  83 

53 

35 

49 

57  to  58 

61 

61 
49  to  58 

36 

34 
41  to  58 

34 

34 

32 


40 
23  to  26 


29  to  32 
.36  to  43 
41  to  52 
38  to  57 
24  to  35 


*  The  results  for  these  timbers  are  deduced  from  experiments  carried  out  by  Bauschinger 
Lanza,  and  others,  on  comparatively  large  specimens. 

THE    BREAKING    WEIGHTS    AND    COEFFICIENTS    OF    BENDING 
STRENGTH    IN    TONS    (of    2240    lhs.)    OF    VARIOUS     RECT- 
ANGULAR   BEAMS,   THE    WEIGHTS    HAVING 
BEEN    UNIFORMLY    DISTRIBUTED. 


Material. 


Yellow  pine  (Quebec), 


1  joists.. 

2  beams. 


Fir  (Baltic),    a  beams 

"         "        II  joists 

"   (Swedish),  a  ioists  

Pine  (Baltic),  2  beams  

Baltic  redwood  deal  (Wyberg),  2  joists  . . . 
Spruce  deals  (St.  John),  3  pairs  with  bridg- 
ing pieces 


Clear 

Span 

Breadth 

between 

in 

Supports 

inches. 

in  inches. 

14a 

3, 

142 

^k 

126 

>4 

126 

14 

142 

3* 

142 

3 

ia6 

•3* 

14a 

3 

14a 

3 

Depth 

in 
inches. 


15 
»4 
II 

9 

'3* 
9 


Mean 
Breaking 
Weight  of 
each  Joist 
or  Beam. 


5.66 

7.89 

60.97 

46.6 

8.29 

5'7 

58.43 

5.75 

6.81 


Coef- 
ficient of 
Bending 
Strength. 


2.48 

t.aS 

1.83 

1.6 

2.0B 

a. 49 

2.24 

2.5a 

a.qB 


THE    BF 
STRE 


Matei 


Yell 


ow  pin 


Pitch  pine. 


Baltic  pine. 


American  ei 


Greenheart. 


Red  pine. 


A'.iS.-The 
ments  carried  i 
pool. 


■Wfl 


TABLES. 


213 


THE    BREAKING    WEIGHTS    AND    COEFFICIENTS   OF   BENDING 
STRENGTH    IN    TONS    (of    2240    lbs.)    OF     VARIOUS    RECT- 
ANGULAR BEAMS  LOADED  AT  THE  CENTRE. 


Material. 


Clear 

Span 
between 
Supports 
in  inches. 


Yellow  pine . 


Pitch  pine. 


Baltic  pine. 


American  elm. 


■Greenheart. 


Red  pine. 


129 
129 
45 
45 
45 
45 
45 
45 
129 
129 
45 
45 
45 
45 
45 
45 
45 
45 
45 
45 
45 
45 
45 
45 
45 
45 
45 
45 
45 
139 
147 
147 


Breadth 
in  inches. 


14 
14 

5 
5 

5 
5 

2i 
2i 

14 

14 

5 
5 
5 
5 
2i 

2J 

5 
S 

2i 

2i 

5 
5 

2i 
2* 

5 

2i 

3 

2i 
2i 

9 
6 
6 


Depth 
in  inches. 


15 

15 

7 

7 

5 

5 

3i 

3i 
15 
15 

7 

7 

7 

7 

■:k 

3i 

7 

7 

3i 

3i 

7 

7 

3i 

34 

5 

7 

3 

3i 

3i 

8 

12 
12 


Breaking 
Weight 
in  tons. 


Coef- 
ficient of 
Bending 
Strength. 


38.15 
34 


9 
7 
I 

05 
•925 
I -075 
59-25 
60.25 
7.8 

9-75 
10.65 
II 

1.6 

1-35 

7 

8.5 

1. 125 

1.2 
14.9 
15-6 

2.65 

2.6 
14 
"•45 

3-85 

4.00 

3-55 
24.5 
7-5 
8.45 


Remarks. 


2.34 
2.og 
1.62 

1-57 

1.67 

1.64 

2.04 

2.37 

3-64 

3-7 

2.14 

2.68 

2.92 

3-03 

3-52 

2.97 

1. 91 

2.34 

2.48 

2.64 

4.1 

4.29 

5-84 

5-73 

7-56 

6.11 

9.62-, 

8.81 

7.82 

8.87 

1.91 

2.15 


Old  timber 


Old  timber 


Old  timber 


A''.5.— The  results  contained  in  the  last  two  tables  are  mainly  deduced  from  experi- 
ments carried  out  under  the  supervision  of  W.  Le  Mesurier,  M.Inst.C.E.,  Dock  Yard,  Liver- 
pool. 


TJ 

m 


■m 


t 


I  \    '  . 


1H 


I      I 


ir 

-•r 
U 


[sl 


ii 


-     r 


H 


fl 


w 


I' it 


r 


II 


IHI 


if 


1 

1 

1 

jU 

■1 

lil 

214  THEORY  OF  STRUCTURES. 

THE  WEIGHTS  AND  CRUSHING   WEIGHTS  OF   ROCKS    ETC. 


Material. 


Weight  per  cu.  ft. 
in  lbs. 


Asphalt 

Basalt,  Scotch 

u    "        Greenstone 

Welsh 

Beton 

Brick,  common 

"      stock  (Eng.) 

"      Sydney,  N.  S 

"      yellow-faced  (Eng.)., 
"       Staffordshire  blue. . . 

"       fire 

"       pressed  (best) 

Brickwork 

Cement,  Portland 

Roman 

Clay 

Concrete,  ordinary 

"         in  cement. .    . . .. 

Earth  

Firestone 

Freestone 

Glass,  flint 

"       crown 

"       common  green 

"       plate 

Granite,  Aberdeen  gray.. . , 

red 

"         Cornish 

Sorrel 

Irish 

U.  S.  (Quincy)..., 

Argyll 

Gneiss 

Limestone , 

Lime,  quick 

Mortar , 

"     (average) 

Masonry,  common  brick.., 

"  in  cement 

rubble 

Marble,  statuary 

"        miscellaneous.... 

OSlite,  Portland  stone 

Bath  stone 

Sand,  quartz 

"       river 

"       pit 

"       fine 

Sandstone,  red  (Eng.) 

"  Derby  grit.  . . 

••  paving  (Eng.). 

"  Scotch 

U.  S 

Shingle 

Slate,  Anglesea 

"      Cornish , 

"      Welsh 

Trap 


156 
184 
181 
172 


100  to  J35 


150 

112 
86  to  94 

100 

119 

119 

137 
77  to  125 

112 

192 
157 
153 
172 
163 

165 
i66 

167 


96  to  175 
154  to  162 

53 

86  to  119 

106 

116  to  144 

170 
168  to  170 

151 

i?3 
177 
"7 
100 

95 
133 
150 

15(J  lo  157 
153  to  155 


83 

179) 
157  - 
180) 
170 


Crushing  Weight  in 
lbs.  per  sq.  in. 


8,300 
17,200 
16,800 

800  to  1,400 
550  to  800 

2,250 

2,200 

1,440 

7,200 

1,700 
10, 200 

1,700  to  6,000' 


460  to  775 


19,600 
3,000  to  3, SCO' 
27,500 
31,000 
31,000 


10,800 

14,000 
12,800 

10,450 
15,000 
10,900 
19,600 
7,500  lo  9,c 


oo- 


120  10  240 

500  to  800 
760 
y^  of  cut  stone- 

3,200 

8,000  to  9,700' 

4,100 


5.700 

3,100 
5,700  to  O.ooo' 
5,300  to  7,800. 

5.300 

10,000 

10 
24,000 


Ti 

M( 
Ml 


Brass 

Bronze.. .. 
Cast-iron,, 
Copper. . . , 

Fir 

Glass 

Gold 

Gun-metai. . 
Iron  wire  . . 

Lead 

Oak 

Platinum... 

Silver 

Steel,  unhardi 

"      hardene 
Tin 

Wrought-iron 
tt 

Zinc,  cast... 
hammer 


'1 


i   kj 


TABLES. 

ACTORS   OF    SAFETY. 

Good  Ordinary  Work. 

Timber 4  to  5  for  dead  load,        8  to  10  for  live  load. 

Metals 3  "       "       "  6 

Masonry 4 8 

EXPANSIONS  OF   SOLIDS. 


21$ 


((      tt 


Materials. 


Brass 

Bronze 

Cast-iron 

Copper 

Fir 

Glass 

Gold 

Gun-metal 

Iron  wire 

Lead 

Oak 

Platinum 

Silver 

Steel,  unhardened 

"     hardened 

Tin 

Wrought-iron  (bar) 

"  (for  smith-work), 
Zinc,  cast 

"     hammered 


Linear  Expansion  per  Unit  of  Length. 


From  3a°  F.  to 
212°  F. 


.00x868 

.00182 

.001075 

,001718 

.00352 

.00861 

.001466 

.00181     = 

.00144 

. 0002848 

. 000746 

.000884 

.001909 

.001079 

.00124 

.002173 

.001235 

.001182 

.00^2941 

.003108 


ffiff 
laVr 

TlVf 

itK 

itVt 
til 


=  Bi 


BtST 
Tfftl 


From  3 


Expansion 
in  Built. 


12°  F.  to        From  32°  F. 
°  F.  to  212°  F. 


572°  F. 


.001883  =  ,fj 


.001468  =  tIt 


.0065 
.0054 
.0033 
.0055 

.0027 


.0057 

.0036 

.0066 

.0036 
.0058 


m 


m 


i'-i' 


w 


<>     ! 


til        ' 
11 


I 


■in 


'     ill 


Hii 


iilil 


i 

1\^ 


• '.'  ; 


'  '.ii 


^^y 


2l6 


THEORY  OF  STRUCTURES. 


Ill 


EXAMPLES. 

r.  How  many  square  inches  are  there  in  the  cross-section  of  an  iron 
rail  weighing  30  lbs.  per  lineal  yard?  How  many  in  a  yellow-pine  beam 
of  the  same  lineal  weight?  Ans.  3  sq.  in.;  45  sq.  in. 

2.  A  vertical  wrought-iron  bar  60  ft.  long  and  i  in.  in  diameter  is 
fixed  at  the  upper  end  and  carries  a  weight  of  2000  lbs.  at  the  lower  end. 
Find  the  factors  of  safety  for  both  ends,  the  ultimate  strength  of  the 
iron  being  50,000  lbs.  per  sq.  in.  Ans.   191^^;  i^fi-^. 

3.  A  vertical  rod  fixed  at  both  tads  ii-  w  jighted  with  a  load  iv  at  an 
intermediate  point.  How  is  the  load  distributed  in  the  tension  of  the 
upper  and  compression  of  the  lowc.  portion  of  the  rod? 

w,    .  Inversely  as  the  lengths. 

4.  Find  the  length  of  a  steel  bar  of  sp.  gr.  7.8  which,  when  suspended 
vertically,  would  break  by  its  own  weight,  the  ultimate  strength  of  the 
metal  being  60,000  lbs.  per  sq.  in.  Ans,  17,723  ft. 

5.  The  iron  composing  the  links  of  a  chain  is  \  in.  in  diameter;  the 
chain  is  broken  under  a  pull  of  10,000  lbs.  What  is  the  corresponding 
tenacity  per  sq.  in.  ?  Ans,  57,272j\  lbs. 

6.  A  vertical  iron  suspension-rod  90  ft.  long  carries  a  load  of  20,000 
lbs.  at  its  lower  end  ;  the  rod  is  made  up  of  three  equal  lengths  square 
in  section.  Find  the  sectional  area  of  each  length,  the  ultimate  tenacity 
of  th  ~  'ron  being  50,000  lbs.  per  sq.  in.,  and  5  a  factor  of  safety. 

Ans.  V^  sq.  in. ;  Vff^  sq.  in.  ;  ^^%^  sq.  in. 

7.  If  the  rod  in  the  previous  question  is  of  a  conical  form,  what 
should  be  the  area  of  the  upper  end  ?  Also  find  the  intensities  of  the 
tension  at  30  and  60  ft.  from  the  lower  end. 

Ans.  2.0407  sq.  in.;  9999.612  lbs.,  9999.605  lbs.  per  sq.  in. 

8.  The  dead  load  of  a  bridge  is  5  tons  and  the  live  load  10  tons  per 
panel,  the  corresponding  factors  of  safety  being  3  and  6.  If  the  two  loads 
are  taken  together,  making  9  tons  per  panel,  what  factor  of  safety  would 
you  use  ?  Ans.  5. 

9.  The  end  of  a  beam  10  in.  broad  rests  on  a  wall  of  masonry.  If  it  be 
loaded  with  10  tons,  what  length  of  bearing  surface  is  necessary,  the  safe 
crushing  stress  for  stone  being  150  lbs.  per  sq.  in.  ?  Ans.  \i\  in. 

10.  Find  diameter  of  bearing  surface  at  the  base  of  a  column  loaded 
with  20  tons,  the  same  stress  being  allowed  as  in  the  preceding  question. 

Ans,   1/380. 1 2. 


II.  In 

alternatel} 
the  chain 
the  metal 


12 

.12  in 


Ar 

und 


13-  A  J 
cable  stret 
respond  inj 

14.  A  1 
1,200,000  1 
under  a  pi 

15.  A 

5oooolbs. 

16.  A  V 
pull  of  52,5 
E  being  21; 

17.  A  A 
between  t 
E  =  29,oo< 
the  tempei 

18.  Wh 
pressure  is 
surface  an< 

19.  A  f 
done  in  sti 

20.  A  r 
subjected 
elongation 

21.  An 
to  a  tensic 
show  by  a 
within  the 

22.  A  t 
30  ft.  from 
what  thicl 
pillar  and  ' 


EXAMPLES. 


217 


11.  In  the  chain  of  a  suspension-bridge  five  flat  links  dovetail  with  four 
alternately,  and  a  cylindrical  pin  passes  through  the  eyes.  The  pull  on 
the  chain  is  200  tons.  Find  the  area  of  the  pin,  the  bearing  strength  of 
the  metal  being  6  tons  per  sq.  in.  Ans.  \\  sq.  in. 

12.  An  iron  bar  of  uniform  section  and  10  ft.  in  length  stretches 
.12  in.  under  a  unit  stress  of  25,000  lbs.     Find  E. 

Ans.  25,000,000  lbs. 

13.  A  ship  at  the  end  of  a  600-ft.  cable  and  one  at  the  end  of  a  500-ft. 
cable  stretch  the  cables  3  in.  and  2\  in.,  respectively.  What  are  the  cor- 
responding strains.'  Ans.  5^5. 

14.  A  rectangular  timber  tie  is  12  in.  deep  and  40  ft.  long.  If  ^  = 
1,200,000  lbs.,  find  the  proper  thickness  of  the  tie  so  that  its  elongation 
under  a  pull  of  270,000  lbs.  may  not  exceed  1.2  in.  Ans.  7^  in. 

15.  A  wrought-iron  bar  60  ft.  long  is  stretched  5  in.  by  a  pull  of 
5ooo^bs.     Find  its  diameter,  E  being  25,000,000  lbs.  Ans.  .59  in. 

16.  A  wrought-iron  rod  984  ft.  long  alternately  exerts  a  thrust  and  a 
pull  of  52,910  lbs. ;  its  cross-section  is  9.3  sq.  in.  Find  the  loss  of  stroke, 
E  being  29,000,000  lbs.  Atis.  4.632  in. 

17.  A  wrought-iron  bar  2  sq.  in.  in  sectional  area  has  its  ends  fixed 
between  two  immovable  blocks  when  the  temperature  is  at  32°  F.  If 
E  =  29,000,000  lbs.,  what  pressure  will  be  exerted  upon  the  blocks  when 
the  temperature  is  100°  F.  ?  Ans.  27388I  lbs. 

18.  What  should  be  the  diameter  of  the  stays  of  a  boiler  in  which  the 
pressure  is  30  lbs.  per  sq.  in.,  allowing  one  stay  to  each  i^  sq.  ft.  of 
surface  and  a  stress  of  3500  lbs.  per  sq.  in.  of  section  of  iron  ? 

Ans.   li  in. 

19.  A  force  of  10  lbs.  stretches  a  spiral  spring  2  in.  Find  the  work 
done  in  stretching  it  successively  i  in.,  2  in.,  3  in.,  up  to  6  in. 

Ans.  |,  V,  V-.  ¥-.  -F.  ^^  in. -lbs. 

20.  A  roof  tie-rod  142  ft.  in  length  and  4  sq.  in.  in  sectional  area  is 
subjected  to  a  stress  of  80,000  lbs.  U  E  =  30,000,000  lbs.,  find  the 
elongation  of  the  rod  and  the  corresponding  work. 

Ans.   1. 136  in.;  3786I  ft.-lbs. 

21.  An  iron  wire  i  in.  in  diameter  and  250  ft.  in  length  is  subjected 
to  a  tension  of  600  lbs.,  the  consequent  strain  being  j^.  Find  E,  and 
show  by  a  diagram  the  amount  of  work  done  in  stretching  the  wire 
within  the  limits  of  elasticity.  Ans.   1 4,661, 8 iS^^  lbs. 

22.  A  timber  pillar  30  ft.  in  length  has  to  support  a  beam  at  a  point 
30  ft.  from  the  ground.  If  the  greatest  safe  strai..  of  the  timber  is  3-J^, 
what  thickness  of  wedge  should  be  driven  between  the  head  of  the 
pillar  and  the  beam  ?  Ans.  -^  ft. 


I-  ,   .« 


-■•■-'» 


..  .ir 


2l8 


THEORY  OF  STRUCTURES. 


23.  An  hydraulic  hoist-rod  50  ft.  in  length  and  i  in.  in  diameter  is 
attached  to  a  plunger  4  in.  in  diameter,  upon  which  the  pressure  is 
800  lbs.  per  sq.  in.  Determine  the  altered  length  of  the  rod,  E  being 
30,000,000  lbs.  Ans.  .0213  ft. 

24.  A  short  cast-iron  post  is  to  sustain  a  thrust  of  1000  lbs.,  the  ul- 
timate crushing  strength  of  the  iron  being  80,000  lbs.  per  sq.  in.  and  ic 
a  factor  of  safety.  Find  the  dimensions  of  the  post,  which  is  rectangular 
in  section  with  the  sides  in  the  ratio  of  2  to  i.  Ans.  4  in.;  2  in. 

25.  The  length  of  a  cast-iron  pillar  is  diminished  from  20  ft.  to  19.97 
ft.  under  a  given  load.  Find  the  strain  and  the  compressive  unit  stress, 
E  being  17,000,000  lbs.  Ans.  .0015  ;  25,500  lbs.  per  sq.  in. 

26.  A  rectangular  timber  strut  24  sq.  in.  in  sectional  area  and  6  ft.  in 
length  is  subjected  to  a  compression  of  14,400  lbs.  Determine  the 
diminution  of  the  length,  E  being  1,200,000  lbs.  Ans.  .003  ft. 

27.  Find  the  height  from  which  a  weight  of  200  lbs.  may  be  dropped 
so  that  the  maximum  admissible  stress  produced  in  a  bar  of  i  sq.  in. 
section  and  5  ft.  long  may  not  exceed  20,000  lbs.  per  sq.  in.,  the  co- 
efficient of  elasticity  being  27,000,000  lbs. 

Ans.  ^j  ft.,  or,  more  accurately,  j^o  ^^• 

28.  Find  the  H.  P.  required  to  raise  a  weight  of  10  tons  up  a  grade 
of  I  in  12  at  a  speed  of  6  miles  per  hour  against  a  resistance  of  9  lbs.  per 
ton.  Ans.  31.3. 

29.  A  square  steel  bar  10  ft.  long  has  one  end  fixed  ;  a  sudden  pull  of 
40,000  lbs.  is  exerted  at  the  other  end.  Find  the  sectional  area  of  the 
bar  consistent  with  the  condition  that  the  strain  is  not  to  exceed  j\^. 
E  =  30,000,000  lbs.     Find  the  resilience  of  the  bar, 

Ans.  2  sq.  in. ;  533^  ft.-lbs. 
_j,Q  30,  How  much  work  is  done  in  subjecting  a  cube  of  125  cu.  in.  uf 
iron  to  a  tensile  stress  of  jooo  lbs.  per  sq.  in.  }  Ans.   11^  ft.-lbs. 

31.  A  signal-wire  2000  ft.  in  length  and  ^  in.  in  diameter  is  subjected 
to  a  steady  stress  of  300  lbs.  The  lever  is  suddenly  pulled  back,  and  the 
corresponding  end  of  the  wire  moves  through  a  distance  of  4  in.  De- 
termine the  instantaneous  increase  of  stress.  Ans.  S^\jI  lbs. 

32.  If  the  total  back-weight  is  350  lbs.,  what  is  the  range  of  the  sig- 


nal end  of  the  wire? 


Ans.  -rifh  ft. 


33.  A  steel  rod  of  length  L  and  sectional  area  A  has  its  upper  end 
fixed  and  hangs  vertically.  The  rod  is  tested  by  means  of  a  ring  weigh- 
ing 60  lbs.  which  slides  along  the  rod  and  is  checked  by  a  collar  screwed 
to  the  lower  end.  A  scale  is  marked  upon  the  rod  with  the  zero  at  the 
fixed  end.  If  the  strain  in  the  steel  is  not  to  exceed  7-J-5,  what  is  the 
reading  from  which  the  weight  is  to  be  dropped  }  What  should  be  the 
reading  of  the  collar  }    E  —  35,000,000  lbs. 

Ans.  Distance  from  point  of  suspension  =  (|§J  —  l^A)L  ;  ^^^L. 


34-  A  io 
pending  roc 
2  sq.  in.,  le 
duced. 

35-  If  th 
2000  lbs.  is  J 

36.  Steal 
upon    a   pis 
length  and 
done  upon  t 

37.  What 
proof  of  .001 

38.  A  boi 
diameter  har 
ft.  of  rope  w; 
block  being  ■ 
lbs.     What  w 

39-  The  b 
it  was  again  1 

40.  What 
rope  is  not  to 

41.  The  st 
versed.     Shov 

42.  A  weig 
weight  is  furth 
to  oscillate, 
equilibrium. 

43-  If  a  spr 
period  of  oscil 

44.  Show  ti 
a  longitudinal 
and  ml  when  * 

45.  A  steel  1 
20,000  lbs.  per 
per  cubic  inch. 


EXAMPLES. 


219 


34.  A  load  of  1000  lbs.  falls  i  in.  before  commencing  to  stretch  a  sus- 
pending rod  by  which  it  is  carried.  If  the  sectional  area  of  the  rod  is 
2  sq.  in.,  length  100  in.,  and  E  =  30,000,000  lbs.,  find  the  stress  pro- 
duced. Ans.   17,828  lbs.  per  sq.  in. 

35.  If  the  rod  carries  a  load  of  5000  lbs.,  and  an  additional  load  of 
2000  lbs.  is  suddenly  applied,  what  is  the  stress  produced  ? 

Ans.  4.500  lbs.  per  sq.  in. 

36.  Steam  at  a  pressure  of  50  lbs.  per  sq.  in.  is  suddenly  admitted 
upon  a  piston  32  in.  in  diameter.  The  steel  piston-rod  is  48  in.  in 
length  and  2  in.  in  diameter,  E  being  35,000,000  lbs.  Find  the  work 
done  upon  the  rod.  Ans.  117.69  ft. -lbs. 

37.  What  should  be  the  pressure  of  admission  to  strain  the  rod  to  a 
proof  of  .001  }  Ans.  68ff  lbs.  per  sq.  in. 

38.  A  boulder-grappler  is  raised  and  lowered  by  a  wire  rope  i  in.  in 
diameter  hanging  in  double  sheaves.  On  one  occasion  a  length  of  1 50 
ft.  of  rope  was  in  operation,  the  distance  from  the  winch  to  the  upper 
block  being  30  ft.  The  grappler  laid  hold  of  a  boulder  weighing  20,000 
lbs.     What  was  the  extension  of  the  rope,  E  being  15,000,000  lbs.  ? 

Ans.  -rhr  ft. 

39.  The  boulder  suddenly  slipped  and  fell  a  distance  of  6  in.  before 
it  was  again  held.     Find  the  maximum  stress  upon  the  rope. 

Ans.  50,452^  lbs.  per  sq.  in. 

40.  What  weight  of  boulder  may  be  lifted  if  the  proof-stress  in  the 
rope  is  not  to  exceed  25,000  lbs.  per  sq.  in.  ol gross  sectional  area? 

Atts.  78,5711  lbs. 

41.  The  steady  thrust  or  pull  upon  a  prismatic  bar  is  suddenly  re- 
versed.    Show  that  its  effect  is  trebled. 

42.  A  weight  W  is  suspended  by  a  spring,  which  it  stretches.  The 
weight  is  further  depressed  i  ft.,  when  it  is  suddenly  released  and  allowed 
to  oscillate.  Find  its  velocity  at  a  distance  x  from  the  position  of 
equilibrium. 


Ans.  y  10(1  —  10.1-^)-"—. 


43.  If  a  spring  deflects  .001  ft.  under  a  load  of  i  lb.,  what  will  be  the 
period  of  oscillation  of  a  weight  of  14  lbs.  upon  the  spring? 

44.  Show  that  the  change  of  a  unit  of  volume  of  a  solid  body  under 

a  longitudinal  stress  is  a(  i ),  which  becomes  —  if  ///  =  4,  as  in  metals, 

\        ml  2 

and  «//when  tn  =  2,  as  in  india-rubber  (page  142). 

45.  A  steel  bar  stretches  TjVif'h  of  its  original  length  under  a  stress  of 
20,000  lbs.  per  sq.  in.  Find  the  change  of  volume  and  the  work  done 
per  cubic  inch.  Ans.  ^■^^\.\\ ;  |  ft. -lb.  per  cu,  in. 


.  I  '  '•■« 


,j 


.1  ? 


1 


t".  1 

'I  ii 
1 
I 


'm 


t 


i[-  ^ 


!  '.      I  ,    » 


220 


THEORY  OF  STRUCTURES. 


m 


iii 


46.  During  the  plastic  deformation  of  a  prismatic  bar,  show  that  the 
change  in  sectional  area  is  proportional  to  the  deformation  calculated  on 
the  altered  length  of  the  bar. 

47.  A  prismatic  bar  of  volume  V  changes  in  length  from  L  to  L  ±  x 
under  the  "  fluid  pressure"/.     Find  the  corresponding  work. 

Ans.  pV\oge{L  ±  x). 

48.  Show  that  the  total  work  done  in  raising  a  number  of  weights 
through  to  a  given  level  is  the  product  of  the  sum  of  the  weights  and  the 
vertical  displacement  of  their  centre  of  gravity. 

49.  An  engine  has  to  raise  4000  lbs.  1000  ft.  in  5  minutes.  What  is  its 
H.  P.  ?     How  long  will  the  engine  take  to  raise  10,000  lbs.  100  ft.  ? 

Ans.  24/j  H.  P. ;  ij  min. 

50.  How  many  men  will  do  the  same  work  as  the  engine  in  the  pre- 
ceding question,  assuming  that  a  man  can  do  900,000  ft. -lbs.  of  work  in 
a  day  of  9  hours  ?  Ans.  480  men. 

51.  Determine  the  H.  P.  which  will  be  required  to  drag  a  heavy  rock 
weighing  10  tons  at  the  rate  of  10  miles  an  hour  on  a  level  road,  the 
coefficient  of  friction  being  0.8.  What  will  be  the  speed  up  a  gradient 
of  I  in  50,  the  same  power  being  exerted  ? 

Ans.  47711 ;  ^l\  miles  per  hour. 

52.  Two  horses  draw  a  load  of  4000  lbs.  up  an  incline  of  i  in  25  and 
1000  ft.  long.     Determine  the  work  done.  Ans.   160,000  ft.-lbs. 

53.  At  what  speed  do  the  horses  walk  if  each  horse  does  16,000  ft.- 
lbs.  of  work  per  minute  ?  Ans.  2r(\  miles  per  hour. 

54.  A  wrought-iron  rod  25  ft.  in  length  and  i  sq.  in.  in  sectional 
area  is  subjected  to  a  steady  stress  of  5000  lbs.  What  amount  of  live 
load  will  instantaneously  elongate  the  rod  by  i  in.,  E  being  30,000,000 
lbs. .'  Ans.  6250  lbs. 

55.  Determine  the  shortest  length  of  a  metal  bar  a  sq.  in.  in  sec- 
tional area  that  will  safely  resist  the  shock  of  a  weight  of  JV  lbs.  falling 
a  distance  of  ^  ft.  Apply  the  result  to  the  case  of  a  steel  bar  i  sq.  in.  in 
sectional  area,  the  weight  being  50  lbs.,  the  distance  16  ft.,  the  proof- 
strain  7-57.  and  E  =  35,000,000  lbs. 

2EJVak 


Ans. 


;.,/  being  the  safe  unit  stress  ;  ^Jlg"  ft. 


ay^-2£lV/ 

56.  A  shock  of  TV  ft.-lbs.  is  safely  borne  by  a  bar  /  ft.  in  length  and  a 
sq.  in.  in  sectional  area.  Determine  the  increased  shock  which  the  bar 
will  bear  when  the  sectional  area  of  the  last  mthoi  its  length  is  increased 

Ans.  Nil 1 . 

\        m      rmj 

57.  The  bar  in  Example  12  is  i  sq.  in.  in  section.  Determine  the  work 
stored  up  in  the  rod  in  foot-pounds  and  compare  it  with  the  work  which 


would  be  st( 
to  4  in. 

58.  If  25 
resilience  fo 

59.  A  st< 
E  =  3S,ooo,c 
area  of  the 
the  weight  o 
determine  tl 
/j  in.,  succei 

An. 

60.  A  hm 
32  F.  Dete 
the  work  stoi 

61.  A  wro 
stretches  .000 
E  =  29,000,0c 

How  may 
ening  walls  tl 

62.  Find  1 
weight  of  a  s 
number  of,  slal 

63.  How  r 
plates,  each  3 
strong  as  the 
wrought-iron  I 

64.  A  horiz 
area  S,  has  its 
/F  suspended 

2ontal.     Show 


f  being  the  inl 
ficient  of  elasti 
65,  A  heav) 
ends  fixed  in  a 
weight.  Find 
throughout). 


EXAMPLES. 


221 


would  be  stored  up  if  for  half  its  length  the  rod  has  its  section  increased 
to  4  in.  Ans.    125  ft.-lbs.;  ^  of  125  ft.-lbs. 

58.  If  25,000  lbs.  per  sq.  in.  is  the  proof-stress,  find  the  modulus  of 
resilience  for  the  i-in.  rod.  Ans.  25  in  in. -lb.  units. 

59.  A  steel  rod  100  ft.  in  length  has  to  bear  a  weight  of  4000  lbs.  If 
E  =  35,000,000  lbs.,  and  if  the  safe  strain  is  .0005,  determine  the  sectional 
area  of  the  rod  (i)  when  the  weight  of  the  rod  is  neglected  ;  (2)  when 
the  weight  of  the  rod  is  taken  into  account.     Also  in  the  former  case, 

determine  the  work  done  in  stretching  tlie  rod  -jV  '"••  tV  '"-  fV  '" 

{^  in.,  successively. 

Ans.  ^g  sq.  in.  ;  //yV  sq-  in. ;  33i,  133^,  300,  ...  1200  in. -lbs. 

60.  A  line  of  rails  is  10  miles  in  length  when  the  temperature  is  at 
32  F.  Determine  the  length  when  the  temperature  is  at  100°  F..  and 
the  work  stored  up  in  the  rails,  E  being  30,000,000  lbs. 

Ans.   10.008  miles  ;  10.24  H.  P. 

61.  A  wrought-iron  bar  25  ft.  in  length  and  i  sq.  in.  in  sectional  area 
stretches  .0001745  ft.  for  each  increase  of  1°  F.  in  the  temperature.  If 
E  =  29,000,000  lbs.,  determine  the  work  done  by  an  increase  of  20°  F. 

How  may  this  property  of  extension  under  heat  be  utilized  in  straight- 
ening walls  that  have  fallen  out  of  plumb  ?  Ans.  7.064  ft.-lbs. 

62.  Find  the  work  done  in  raising  a  Venetian  blind,  iv  being  the 

weight  of  a  slat,  a  the  distance  between  consecutive  slats,  and  n  the 

number  of  slats.  ,  n{n  +  i) 

Ans.    wa . 

63.  How  many  |-in.  rivets  must  be  used  to  join  two  wrought-iron 
plates,  each  36  in.  wide  and  \  in.  thick,  so  that  the  rivets  may  be  as 
strong  as  the  riveted  plates,  the  tensile  and  shearing  strength  of 
wrought-iron  being  in  the  ratio  of  10  to  9?  Atis.  17  rivets  (16.3). 

64.  A  horizontal  string,  without  weight,  of  length  2a  and  sectional 
area  S,  has  its  two  ends  fixed  in  the  same  horizontal  plane.  A  weight 
/K  suspended  from  its  centre  draws  the  string  slightly  out  of  the  hori- 
zontal.    Show  that,  approximately. 


'=K^')*  '""*  '^^'^l^)* 


m 


t  being  the  intensity  of  the  tension,  d  the  depression,  and  E  the  coet- 
ficient  of  elasticity. 

65.  A  heavy  wire  of  length  2a,  sectional  area  5,  and  weight  W  has  its 
ends  fixed  in  a  horizontal  plane  and  is  allowed  to  deflect  under  its  own 
weight.  Find  the  deflection  d  and  the  tenacity  /  (assumed  uniform 
throughout). 


•f!s 


222 


THEORY  OF  STRUCTURES. 


\\i\ 


•> 


66.  A  length  270  ft.  of  wire  i  sq.  in.  in  section  and  of  sp.  gr.  7.8  is 
subjected  to  tlie  above  conditions.  Find  the  tenacity  of  the  wire  and  the 
deflection,  the  coefficient  of  elasticity,  E,  being  25,300,000  lbs. 

67.  A  brick  wall  2  ft.  thick,  12  ft.  high,  and  weighing  112  lbs.  per 
cu.  ft.  is  supported  upon  solid  pitch-pine  columns  9  in.  in  diameter. 
10  ft.  in  length,  and  spaced  12  ft.  centre  to  centre.  Find  the  compress- 
ive unit  stress  in  the  columns  (1)  at  the  head;  (2)  at  the  base.  The  tim- 
ber weighs  50  lbs.  per  cu.  ft.  Ans.  507.03  lbs. ;  510.5  lbs. 

68.  If  the  crushing  stress  of  pitch-pine  is  5300  lbs.  per  sq.  in.  and  the 
factor  of  safety  10,  find  the  height  to  which  the  wall  may  be  built. 

Ans.  12.46  ft. 

69.  Determine  the  diameter  of  the  wrought-iron  columns  which 
might  be  substituted  for  the  timber  columns  in  question  67,  allowing  a 
working  stress  in  the  metal  of  7500  lbs.  per  sq.  in.  Ans.  2.36  in. 

70.  Find  the  greatest  length  of  an  iron  suspension-ro^i  which  will  carry 
its  own  weight,  the  stress  being  limited  to  4  tons  per  sq.  in.  What  will 
be  the  extension  under  this  load,  E  being  12,500  tons.' 

Ans.  2700  ft. ;  .864  ft, 

71.  A  horizontal  cast-iron  bar  i  ft.  long  exactly  fits  between  two  verti- 
cal plates  of  iron.  How  much  should  its  temperature  be  raised  so  that 
it  might  remain  supported  between  the  plates  by  the  friction,  the  coef- 
ficient of  friction  being  j^  ?  Ans.  -^''  F. 

72.  The  fly-wheel  of  a  40  H.  P.  engine,  making  50  revolutions  per 
minute,  is  20  ft.  in  diameter  and  weighs  12,000  lbs.  What  is  its  kinetic 
energy? 

If  the  wheel  gives  out  work  equivalent  to  that  done  in  raising  5000 
lbs.  through  a  height  of  4  ft.,  how  much  velocity  does  it  lose? 

Tho  axle  of  the  fly-wheel  is  12  in.  in  diameter.  What  proportion  of 
the  H.  P.  is  required  to  turn  the  wheel,  the  coefficient  of  friction  being 
.08? 

If  the  fly-wheel  is  disconnected  from  the  engine  when  it  is  making 
50  revolutions  per  minute,  how  many  revolutions  will  it  make  before  it 
comes  to  rest  ? 

Ans.  51 1,260.4  ft.-lbs. ;  1.04  ft.  per  sec. ;  |ths  ;  169.4. 

73.  The  velocity  of  flow  of  water  in  service-pipe  48  ft.  long  is  64  ft. 
per  sec.  If  the  stop-valve  is  closed  in  \  of  a  sec,  find  the  increase  of 
pressure  near  the  valve.  Ans.  375  lbs.  per  sq.  in. 

74.  Work  equivalent  to  50  ft.-lbs.  is  done  upon  a  bar  of  constant 
sectional  area,  and  produces  in  it  a  uniform  tensile  stress  of  10,000  lbs. 
per  sq.  in.     Find  the  cubic  content  of  the  bar,  E  being  30,000,000. 

Ans.  360  cu.  in. 

75.  A  fly-wheel  weighs  20  tons  and  its  radius  of  gyration  is  5  ft.    How 


much  H 
per  min 

76.  '. 
30,000  fi 
was  con 
same  six 

n.  / 

uniform 
fulfil  ? 

78.  E 

1,650,000 
nules  pel 

79-  A 
miles  an 
weight  ol 

80.  A 
long,  agai 
H.  P.  of  1 
H.  P.,  is  . 

81.  Th 
tional  are 
metal  is  n 
pillar  mig 

82.  A 
area  carrit 
termine  tl 
35,000.000 

83.  A 
strained  to 
the  rod,  E 

84.  Wh 
nail  is  driv 
of  the  mass 

85.  A  h 
10  ft.  per  s( 
force  on  th 

86.  A  hi 
a  nail  5  in. 
the  rnomen 
steady  press 


•\ir  n 


"WZi 


m 


EXAMPLES. 


223 


much  work  is  given  out  while  the  speed  fails  from  60  to  50  revolutiono 
per  minute  ?  ^hts.  y+iVA  ft. -tons 

76.  The  resilience  of  an  iron  bar  i  sq.  in.  in  section  and  20  ft.  long  is 
30,000  ft.-l'  Vhat  would  be  the  resilience  if  for  19  ft.  of  its  length  it 
was  compc  of  iron  2  sq.  in.  in  section,  the  remaining  foot  being  the 
same  size  as  before  ?  Ans.  8625  ft.-lbs. 

77.  A  particle  under  the  action  of  a  number  of  forces  moves  with  a 
uniform  velocity  in  a  straight  line.  What  condition  must  the  forces 
fulfil  }  Ans.   Equilibrium. 

78.  Determine  the  constant  effort  exerted  by  a  horse  which  does 
1,650,000  ft.-lbs.  ol  work  in  one  hi)ur  when  walking  at  the  rate  of  2^ 
miles  [ler  hour.  Ans.  125  lbs. 

79.  A  train  is  drawn  by  a  locomotive  of  160  H.  P.  at  the  rate  of  60 
miles  an  hour  against  a  resistance  of  20  lbs.  per  ton.  What  is  tlie  gross 
weight  of  the  train  ?  Av.s.   50  tons. 

80.  A  train  of  292J  tons  is  drawn  up  an  incline  of  i  in  75,  5^  miles 
long,  against  a  resistance  of  10  lbs.  per  ton,  in  ten  minutes.  Find  the 
H.  P.  of  the  '  'ine.  The  speed  on  the  level,  the  engine  exerting  769.42 
H.  P.,  is  43.  "S  per  hour.    Wiiat  is  the  resistance  in  pounds  per  ton  ? 

Ans.   1027  H.  P. ;  22.7  lbs.  per  ton. 

81.  The  dead  load  upon  a  short  hollow  cast-iron  pillar  with  a  sec- 
tional area  of  20  sq.  in.  is  50  tons  (of  2000  lbs.).  If  the  strain  in  the 
metal  is  not  to  exceed  .0015,  find  the  greatest  live  load  to  which  the 
pillar  might  be  subjected,  E  being  17,000,000  lbs.         Ans.  255,000  lbs. 

82.  A  steel  suspension-rod  30  ft.  in  length  and  ^  sq.  in.  in  sectional 
area  carries  3500  lbs.  of  the  roadway  and  3000  lbs.  of  the  live  load.  De- 
termine the  gross  load   and  also  the  extension  of   the  rod,  E  being 


',5,000.000  lbs. 


^i"^-  ^Ih  ft- 


83.  A  steel  rod  10  ft.  in  length  and  ^  sq.  in.  in  sectional  area  is 
strained  to  the  proof  by  a  tension  of  25,  00  lbs.  Find  the  resilience  of 
the  rod,  E  being  35,000,000  lbs.  Ans.   178^  ft.-lbs. 

84.  What  form  does  the  useful  work  done  by  a  hammer  take  when  a 
nail  is  driven  into  any  material  ?  What  becomes  of  the  rest  of  the  energy 
of  the  mass  of  the  hammer  after  striking  the  blow  } 

85.  A  hammer  weighing  2  lbs.  strikes  a  steel  plate  with  a  velocity  of 
10  ft.  per  sec,  and  is  brought  to  rest  in  .0001  sec.  What  is  the  average 
force  on  the  steet.'  Ans.  6250  lbs. 

86.  A  hammer  weighing  10  lbs.  strikes  a  blow  of  10  ft.-lbs.  and  drives 
a  nail  5  in.  into  a  piece  of  timber.  Find  the  velocity  of  the  hammer  at 
the  moment  of  contact,  and  the  mean  resistance  to  entr3\  Also  find  the 
steady  pressure  that  will  produce  the  same  effect  as  the  hammer. 

Ans.  8  ft.  per  sec. ;  240  lbs. ;  480  lbs. 


:W 


5< 


M 


224 


THEORY  OF  STRUCTURES. 


87.  When  :i  nail  is  driven  into  wood,  why  do  the  blows  seem  to  have 
little  if  any  effect  unless  the  wood  is  backed  up  by  a  piece  of  metal  or 
stone  ? 

88.  In  Question  86,  takinj^  the  weight  of  the  nail  to  be  4  oz.  and  the 
weight  of  tiie  piece  of  timber  to  be  100  lbs.,  find  the  depth  and  time  ol 
the  penetration  (ii)  when  the  timber  is  fixed ;  (b)  when  the  timber  is  free 
to  move. 

Also  in  case  {p)  find  tlie  distance  through  which  the  timber  moves. 
Ans. — (d)  j','  in.;  b'b  sec. 

(/')  .44245  ill.;  .0009448  sec.  ;  .04113  in. 

Si).  Show  that  the  greater  part  of  the  eiicigy  of  impact  is  expended 
in  local  damage  at  high  velocities,  and  in  straining  the  impinging  bodies 
a.s  a  whole  at  low  velocities. 

90.  A  pile-driver  of  300  lbs.  falls  20  ft.,  and  is  stopped  in  ^  sec.  What 
is  the  average  force  exerted  on  the  pile  ?  Ans,  3344  lbs. 

yi.  A  weight  falls  16  ft.  and  docs  2560  ft.-lbs.  of  work  upon  a  pile 
which  it  drives  4  in.  against  a  uniform  resistance.  Find  the  weight  of 
the  ram,  and  the  resistance.  Ans.  160  lbs. ;  7680  li)s. 

92.  A  pitch-pine  pile  14  in.  square  is  20  ft.  above  ground,  and  is 
being  driven  by  a  falling  weight  of  112  lbs.  \{  E  =  1,500,000  lbs.,  find 
the  fall  so  that  the  inch-stress  at  the  head  of  the  pile  may  be  less  than 
800  11)S. 

Supposing  that  the  pile  sinks  2  in.  into  the  ground,  by  how  much 
would  it  be  safe  to  increase  the  fall.' 

Ans.  7.456  ft. ;  1 16.5  ft. 

93.  A  weight  of  W\  tons  falls  //  ft.,  and  by  n  successive  instantaneous 
blows  drives  an  inelastic  pile  weighing  \Vi  tons  a  ft.  into  the  ground. 
Assuming  the  pile  and  weight  to  be  inelastic,  lind  (</)  the  mean  ellectiv:> 
resistance  of  tlie  ground. 

If  the  ground-resistance  increases  directly  as  the  depth  of  penetration, 
find  (/')  how  far  the  pile  will  sink  under  tlie  /th  blow.  If  the  head  of  tin* 
pile  is  crushed  for  a  length  of  .1  ft.,  x  being  very  small  as  compared  with 

the  depth  —  of  penetration,  find  (i)  the  mean  thrust,  during  the  blow. 

between  the  weight  and  hammer ;  (2)  the  time  of  penetrating  the  ground  : 
(3)  the  time  during  which  the  blow  arts. 


Ans,-(ii) 


ri) 


w 


«// 


•  '  1 


(2) 


\Vx  4-  \W      <t 


\\\ 


(3) 


4«  \'h  '     "    4  j/A 
94.   An   inelastic  pile  weighing  788  lbs.  is  driven   3i  feet  into  the 
ground  by  120  blows  from  a  weight  of  112  lbs.  falling  30  ft.     Find  the 


steady  loa 
tin;  groun 
jienetratit 
nient  of  t 
pile  the  fi 
71O8  lbs.? 
A 

95.  A  s 
an  hour  co 
knots  an  h 
of  collision 

y6.  A  h; 
|Ji;r  sec,  dr 
nail? 

97.  A  be 
what  lieigiit 
the  effect  of 

98.  A  rifl 
'Icr  wci;,'lis  S 
<»f  the  rifle  i.*^ 
|)owder.  If 
air,  find  thei 
sinks  i  in. 

If  Iheio  i 
111  lizzie- vehjc 

99-  A  leal 
tension  is  in 
taken  at  60  11 

100.   Find 
6  in.  lonjr  ani 
iIk!  axis  of  tl 
How  would  y( 

101.  The  p 
1 1 0111  12.8  ton;- 
tlie  metal,  15 
of  the  bar,  3  bt 

102.  The  SI 
from  a  tensio 
primitive  stren 


EXAMPLES, 


225 


steady  load  upon  the  pile  which  will  produce  the  s;itiu;  cdcct,  assumiiij; 
the  f^round-resistance  to  be  {a)  uniform  ;  {(>)  proportional  to  tlie  dcptii  ol 
jieiietration.  If  tiie  resistance  is  uniform,  how  lon^  (t)  docs  cacii  move- 
iiuMit  of  the  pile  last?  How  many  blows  (d)  are  re(|iiirefl  to  drive  tiie 
l)ilc  the  first  lialf  of  tiic  depth,  viz.,  \\  ft.,  tiie  ground-resistance  iieing 
7168  lbs.  ?    How  far  {c)  does  the  pile  sink  under  the  last  IjIow  } 

Alls,    (a)    14,336  lbs. ;    {/>)    28,672  lbs.  ;    (c)  .0107  sec.  ;    (d)  30; 
(c)  .016  in. 

95.  A  steamer  of  8000  tons  displacement  sailing  due  east  at  16  knots 
an  hour  collides  witii  a  steamer  of  5000  tons  displacement  sailing  at  10 
knots  an  hour.  Find  the  energy  of  collision  if  the  latter  at  the  moment 
of  collision  is  going  (1)  due  west;  (2)  north-west;  (3)  north-east. 

t)G.  A  hammer  weighing  2  lbs.  strikes  a  nail  with  a  velocity  of  15  ft. 
per  sec,  driving  it  in  A  in.  What  is  the  mean  jjressure  overcome  tjy  the 
nail  ?  A/IS.  673  iljs. 

97.  A  beam  will  safely  carry  i  ton  witli  a  deflection  of  1  in.  Fr(jm 
what  height  may  a  weight  of  100  lbs.  drop  without  injuring  it,  neglecting 
the  elfect  of  inertia  ?  //w.   11.2  in. 

98.  A  rifie-l)iillet  .45  in.  in  diameter  weighs  1  oz. ;  the  cliarge  of  pow- 
der wei^lis  85  grains;  the  mnzzle-velocity  is  1350ft.  per  sec;  the  weight 
of  the  rifle  is  9  lbs.  Nc}:;lcctin}r  the  twist  determine  the  energy  of  1  lb.  of 
powder.  If  the  bullet  loses  \  of  its  velocity  in  its  passage  tlirougii  the 
air,  find  the  average  force  of  llu;  blow  on  the  target  into  wiiich  tiie  bullet 
binks  I,  in. 

If  tliere  is  a  twist  of  i  in  20  in.,  find  the  charge  to  give  the  same 
muzzle- velocity,  the  length  of  tl;':  barrel  being  33  in. 

99.  A  leather  belt  runs  at  2400  ft.  per  minute.  Find  how  much  its 
tension  is  increased  by  centrifugal  action,  the  weight  of  leather  being 
taken  at  60  lbs,  per  cubic  foot,  ,  Ans.  20%  lbs. 

100.  Find  the  centrifugal  force  arising  from  a  cylindrical  crank  pin 
6  in.  long  and  3^  in.  in  diameter,  the  axis  of  the  pin  being  12  in.  fnmi 
the  axis  of  the  engine-shaft,  \.!Mch  makes  200  revolutions  per  minute. 
How  would  you  balance  such  a  pin  }  ^Ins.  55.02  lbs. 

101.  The  i)ull  on  one  of  the  tension-bars  of  a  lattice  girder  fluctuates 
Irom  12.8  tons  to  4  tons.  If  24  tons  is  the  statical  breaking  strength  ot 
ilie  metal,  15  tons  the  primitive  strength,  determine  the  sectional  area 
of  the  bar,  3  being  a  factor  of  safety.        Ans.  2.15  sq.  in.  (Launhardt) ; 

1.87  sq.  in.  (Unwin). 

102.  The  stress  in  a  diagonal  of  a  steel  bowstring  girder  fluctuates 
from  a  tension  of  15.15  ^""s  to  a  compression  of  7.65  tons.  If  the 
primitive  strength  of  the  metal  is  24  tons  and  the  vibration  strength  12 


H 


I; 


rl 


226 


THEORY  OF  STRUCTURES. 


tons,  find  the  proper  sectional  area  of  the  diagonal,  3  being  a  factor  of 
safety.        Ans.  2.53  sq.  in.  (Weyrauch) ; 

1.7  sq.  in.  (Unvvin),  40  tons  per  sq.  in.  being  statical 
strength. 

103.  A  wrought-iron  screw-shaft  is  driven  by  a  pair  of  cranks  set  at 
light  angles.  Neglecting  the  obliquity  of  the  connecting-rods,  and 
assuming  that  the  pull  on  the  crank-pin  is  constant,  compare  the  coef- 
ficients of  strength  (ti'  and  /)  to  be  used  in  calculating  the  diameter  of 
I  he  shaft.     How  is  the  result  affected  by  the  stopping  of  the  engine.'' 

Ans.  a  =  .82/;  <?'  =  |A 

104.  Taking/ =  EX.  as  the  ordinary  analytical  expression  of  Hooke's 
Law,  find  the  value  of  the  modulus  of  elasticity  when  calculated  (i)  from 
the  actual  stress  and  the  elongation  per  unit  of  initial  length  ;  (2)  from 
the  actual  stress  and  the  elongation  per  unit  of  stretched  length. 

Ans.  {\)  E+f;  (2)  E  -f /(i  -1-  A)'  =  E  ^- f{i  -|-  2/I),  if  A  is  small. 

105.  In  a  fly-wheel  weighing  12,000  lbs.  and  making  50  revolutions 
per  minute,  the  centre  of  gravity  is  one  seventeenth,  of  an  inch  out  of  the 
centre.     Find  the  centrifugal  force.  Ans.  50.4  lbs. 

106.  In  the  preceding  question,  if  the  axis  of  rotation  is  inclined  to 
the  plane  of  the  wheel  at  an  angle  cot  ~ '.001,  find  the  centrifugal 
couple,  the  radius  of  gyration  being  10  ft.  Ans.    1028.9  ft.-lbs. 

107.  A  cylinder  and  a  ball  each  of  radius  R  start  from  rest  and  roll 
down  an  inclined  plane  without  slipping.  If  V  is  the  velocity  of  trans- 
lation after  descending  through  a  vertical  distance  N,  show  that 

V^  =  l{2^/i)  in  the  case  of  the  cylinder, 
and 

F"  =  ^{2gh)  in  the  case  of  the  ball. 

:o8.  A  wheel  having  an  initial  velocity  of  10  ft.  per  sec.  ascends  an 
incline  of  i  in  100.  How  far  will  the  wheel  run  along  the  incline,  neg- 
lecting friction  .''  Ans.  232.9  ft. 

IC9.  A  wrought-iron  fly-wheel  10  ft.  in  diameter  makes  63  revolutions 
per  minute.  Find  the  intensity  of  stress  on  a  transverse  section  of  the 
rim,  disregarding  the  influence  of  the  arms.  If  the  wheel,  which  weiglis 
IV  lbs.,  gives  out  work  equivalent  to  that  done  in  raising  W  through  a 
height  of  sJ  ft.  in  i  sec,  what  velocity  will  it  lose  ?  If  the  axle  of  the  wheel 
is  10  in.  in  diameter  and  if  .08  is  the  coefficient  of  friction,  show  that  it 
IV 


will  take 


H.  P.  to  turn  the  wheel. 


2500 

Ans.  16,335  lbs.  ;  3.6  ft.  per  sec. 

no.  If  the  earth  be  assumed  to  be  spherical,  how  much  heat  would 
be  developed  if  its  axial  rotation  were  suddenly  stopped,  a  unit  of  heat 
correspond ing  to  778  ft.-lbs.  ? 


Wcij 
=  Sooo 

in. 

plane,  tl: 

work  wil 

If  the 

tion  in  4 

112.  J 
end  secut 
to  the  otl 
the  anchc 


113.  A 
worked  fr, 
Tlie  wagoi 
and,  after  ( 
•stoppage  o 

"4.  A< 
attached  to 
of  nlV  lbs 
till;  first.     ] 


"5.  A  ( 

\vhich  it  is 
tive  sectior 
-o  ft.  per  se 
find  the  gre; 

116.  {a) 
an  liour  agai 

(^>)  VVith 
in  100.5 

(<■■)  If  tlu- 

(0  ON  the  in( 

(<i)  If   th, 

carriages  (loc 

'iiediately  thi 

-md  the  coefli 


EXAMPLES. 


227 


Weight  of  mass  of  earth 
=  Sooo  miles. 


10"  X  6.029  tons;    diameter  of  earth 


111.  A  body  weighing  50  lbs.  is  projected  along  a  rough  horizontal 
plane,  the  velocity  of  projection  being  100  ft.  per  sec.  What  amount  of 
work  will  have  been  expended  when  the  body  comes  to  rest  ? 

If  the  coefficient  of  friction  is  i,  how  much  work  is  done  against  fric- 
tion in  4  sees.,  and  in  what  time  will  the  body  come  to  rest  } 

Alls.  7763.9  ft.-lbs.  ;  2298I  ft.-lbs;  2\\i\  sees. 

112.  A  chain  /  ft.  in  length  and  a  sq.  in.  in  sectional  area  h;is  one 
end  securely  anchored,  and  suddenly  checks  a  weight  of  VV  lbs.  attaclied 
to  the  other  end,  and  moving  with  a  velocity  of  V  ft.  per  sec.  away  fro  .■ 
the  anchorage.     Find  the  greatest  pull  upon  the  chain. 


Ans.  Pull 


113.  Apply  this  result  to  the  case  of  a  wagon  weighing  4  tons  and 
worked  from  a  stationary  engine  by  a  rope  3  sq.  in.  in  sectional  area. 
The  wagon  is  running  down  an  incline  at  the  rate  of  4  miles  an  hour, 
and,  after  600  ft.  of  rope  have  been  paid  out,  is  suddenly  checked  by  the 
stoppage  or  reversal  of  the  engine  {E  =  15,000,000  lbs.), 

Atts.  26,884  'bs. 

I  [4.  A  chain  /  ft.  in  length  and  a  sq.  in.  in  sectional  area  has  one  end 
attached  to  a  weight  of  W  lbs.  at  rest,  and  at  the  other  end  is  a  weight 
of  n\V  lbs.  moving  with  a  velocity  of  V  ft.  per  second  and  away  from 
the  tlrst.     Find  the  greatest  pull  on  the  chain. 


Alts.  Pull  ^  V 


../  aEWn 
^  k{n  +  I)" 


115.  A  dead  weight  of  10  tons  is  to  act  as  a  drag  upon  a  ship  to 
which  it  is  attached  by  a  wire  rope  150  ft.  in  length  and  having  an  effec- 
tive sectional  area  of  8  sq.  in.  If  the  velocity  of  the  floating  ship  is 
20  ft.  per  second,  and  if  its  inertia  is  equivalent  to  a  mass  of  390  tons, 
liad  the  greatest  pull  on  the  chain  {E  =  15,000,000  lbs.). 

A  us,  208  tons. 

116.  (<i)  A  train  weighing  160  tons  (of  2240  lbs.)  travels  at  30  miles 
an  hour  against  a  resistance  of  10  lbs.  per  ton.     What  H.  P.  is  exerted  } 

(/>)  With  the  same.  H.  P.  what  will  be  the  speed  up  a  gradient  of  i 
in  100? 

(c)  If  the  steam  is  shut  ofT,  how  far  will  the  train  run  before  Stopping 
(i)  on  the  incline;  (2)  on  the  level.'' 

((/)  If  the  draw-bar  suddenly  breaks,  in  what  distance  would  the 
carriages  (100  tons  in  weight)  be  stopped  if  the  brakes  are  applied  im- 
mediately the  fracture  occurs,  the  weight  of  the  brake-van  being  20  tons 
and  the  coefficient  of  friction  .2  .^ 


■H  I' 


'ii4 

'4 

1 

■ 

\:l    '•-♦ 


•'t-   -If 


, 

'■i 


Mm 

,udMm 

228 


THEORY  OF  STRUCTURES. 


(e)  If  the  engine  (weight  =  60  tons)  continued  to  exert  the  same 
power  after  the  fracture,  what  would  be  its  ultimate  speed  ? 

(/)  What  resistance  would  be  required  to  stop  the  whole  train  after 
steam  is  shut  off,  in  1000  yards  on  the  level  ? 

Ans.  {a)  128;  (d)  g^j\  miles  per  hour;  (c)  (i)  199.2  ft., 
(2)  6776  ft.  ;  (flf)  680.3  ft.  on  the  level,  52.9  ft. 
on  the  incline;  (e)  80  miles  an  hour  on  the 
level,  24.6  miles  on  the  incline  ;  (/)  22. 5S  lbs. 
per  ton. 

117.  A  4-in.  X  3-in.  diameter  crank-pin  is  to  be  balanced  by  two 
weights  on  the  same  side  of  the  crank  ;  the  length  of  the  crank  is  12  in. ; 
the  engine  makes  100  revolutions  per  minute;  the  distance  of  the  C.  of 
G  of  each  weight  from  the  axis  of  the  shaft  is  6  in.     Find  the  weights. 

118.  A  shaft  is  worked  with  cranks  at  120°.  Assuming  the  pressure 
on  the  crank-pin  to  be  horizontal  and  constant  in  amcjunt,  compare  the 
coelBcients  of  actual  and  ultimate  strength  to  be  used  in  calculating  the 
diameter  of  the  shaft.  Ans.  a'  =  .507/. 

119.  In  a  horizontal  marine  engine  with  two  cranks  at  riglit  angles 
distant  8  ft.  from  one  another,  weight  of  reciprocating  parts  attached  to 
each  crank  is  lo  tons,  revolutions  75  per  minute,  stroke  4  ft.  Find  the 
alternating  force  and  couple  due  to  inertia. 

Afis.  54.2  tons;  216.8  ft.-tons. 

120.  An  inside-cylinder  locomotive  is  running  at  50  miles  an  hour; 
the  driving-wheels  are  6  ft.  in  diameter;  the  distance  between  the  centre- 
lines of  the  cylinders  is  30  in.,  the  stroke  24  in,,  the  weight  of  one  piston 
and  rod  300  lbs.,  and  the  horizontal  distance  between  the  balance- 
weights  4^  ft. ;  the  diameter  of  the  weight-circle  is  4^  ft.  Find  the 
alternating  force  and  couple,  and  also  the  magnitude  and  position  of 
suitable  balance- weights. 

Ans.  7871  lbs;  9839  ft.-lbs.  ;  106.5  lbs.  ;  27f°. 

121.  The  pressure  equivalent  to  the  weight  of  the  reciprocating  parts 
of  an  engine  is  3  lbs.  per  sq.  in.;  the  stroke  is  36  in.  ;  the  number  of 
revolutions  per  minute  is  45;  the  back-pressure  is  2  lbs.  per  sq.  in.;  ihe 
absolute  initial  steam- pressure  is  60  lbs.  per  sq.  in.;  the  rale  of  expansion 
is  3.  Find  the  pressure  necessary  to  start  the  piston,  and  also  the  effec- 
tive pressure  at  each  ^  of  the  stroke. 

122.  An  engine  with  a  24-in.  cylinder  and  a  connecting-rod  =  six  cranks 
=  6  ft.,  makes  60  revolutions  per  minute.  Show  that  the  pressure  re- 
quired to  start  and  stop  the  engine  at  the  dead-points  =  ^^  of  the  weight 
of  reciprocating  parts. 

123.  Find  the  ratio  of  hrust  at  cross-head  to  tangential  effort  on 
crank-pin  when  the  crank  is  45°  from  the  line  of  stroke,  the  connecting- 
rod  being  =  four  cranks. 


if 


EXAMPLES. 


229 


124.  Draw  the  linear  diagram  of  cranic-effort  in  tlie  case  of  single 
crank,  the  connecting-rod  being  =  four  cranks.  Assume  the  resistance 
uniform  and  a  constant  pressure  of  9000  lbs.  on  the  piston,  the  stroke 
oeing  4  ft.  and  the  number  of  revolutions  per  minute  55.  Also  find  the 
fluctuation  of  energy  in  ft.-lbs.  for  one  revolution. 

125.  An  engine  with  a  connecting-rod  =  six  cranks  =  6  ft.  receives 
steam  at  70  lbs.  pressure  per  sq.  in.,  and  cuts  oft  at  one-quarter  stroke. 
Find  the  crank-effort  when  the  piston  has  travelled  one  third  of  its  for- 
ward stroke.  Diameter  of  piston  =  2  ft.  Also  find  the  position  of  the 
piston  where  its  velocity  is  a  maximum. 

126.  Data:  Stroke  =  3  ft.  ;  number  of  revolutions  per  minute  =  60; 
cut-off  at  one-half  stroke;  initial  pressure  =  56  lbs.  persq.  in.  absolute; 
diameter  Oi  piston  =  10  in. ;  weight  of  reciprocating  parts  =  550  lbs. ; 
back-pressur.^  =  \\  lbs.  per  sq.  in.  absolute.  Find  the  effective  pressure 
at  each  fourtl:  of  the  stroke,  taking  account  of  the  inertia  of  the  piston. 
Also  find  the  pressure  equivalent  to  inertia  at  commencement  of 
stroke. 

127.  A  pair  of  250  H.  P.  engines,  with  cranks  at  90°,  and  working 
against  a  uniform  resistance  and  under  a  uniform  steam-pressure,  are 
running  at  60  revolutions  per  minute.  Assuming  an  indefinitely  long 
connecting-rod,  find  the  maximum  and  minimum  moments  of  crank- 
effort,  the  fluctuation  of  energy,  and  the  coefficient  of  energy. 

128.  An  inside-cylinder  locomotive  runs  at  25  miles  per  hour;  its 
drivers  are  60  in.  in  diameter;  the  stroke  is  24  in. ;  the  distance  between 
the  centre-lines  of  the  cylinders  =  30  in.;  weight  of  reciprocating 
parts  =  500  lbs. ;  horizontal  distance  between  balance-weights  =  59  in. ; 
diameter  of  weight-circle  =  42  in.  Find  the  alternating  force,  alternat- 
ing couple,  and  the  magnitude  and  position  of  suitable  balance-weights. 

Ans.  226.8  lbs.;  41 13.8  ft.-lbs. ;  </>  =  26^ 

129.  Draw  a  diagram  of  crank-effort  for  a  single  crank,  the  connect- 
ing-rod being  equal  to  four  cranks,  the  stroke  4  ft.,  and  the  number  of 
revolutions  per  minute  55.  Assume  a  uniform  resistance  and  a  constant 
pressure  of  9000  lbs.  on  the  piston. 

130.  A  vertical  prismatic  bar  of  weight  ffi ,  sectional  area  .^,  and 
length  L  has  its  upper  end  fixed,  and  carries  a  weight  W%  at  the  lower 
end.     Find  the  amount  and  work  of  the  elongation. 

ea\-  +  ^^^j  •'  "'^^'^  =  ea[-J 

131.  A  right  cone  of  weight  W  and  height  h  rests  upon  its  base  of 
radius  r.     Find  the  amount  and  work  of  the  compression. 

Wh  I   W^ 


Am.  Ext. 


+  lVtlV,  +  lVi^] 


Ans,  Comp.  =  — 


27TEr' 


work  = 


8  rrE'r 


132.  A  tower  of  height  A,  in  the  form  of  a  solid  of  revolution  about  a 
vertical  axis,  carries  a  given  surc/iartfc.     If  the  specific  weight  of  the 


n 


fm  '■ 


m 


■?>'!' 


'U 


T 


i- 


V^ 


230 


THEORY  OF  STRUCTURES. 


material  of  the  tower  is  w,  and  the  radius  of  the  base  a,  determine  the 
curve  of  the  generating  line  so  that  the  stress  at  every  point  of  the  tower 
may  be/.  If  the  surcharge  is  zero  and  the  heiglit  of  the  tower  becomes 
infinite,  show  that  its  volume  remains  finite. 


Ans.  y  =  ac 


wx 


vol.  of  tower  of  infinite  height  i=—na 

w 


VIX 

T. 


133.  Determine  the  generating  curve  when  the  tower  in  the  last 
question  is  hollow,  the  hollow  part  being  in  the  form  of  a  right  cylinder 
upon  a  circular  base  of  given  radius  R. 

Ans.  y  —  R*  -  (rt"  -  R')e 

134.  A  iieavy  vertical  bar  of  length  /  and  specific  weight  w  is  fixed 
at  its  upper  end  and  carries  a  given  weight  IV  at  the  lower  end.  Deter- 
mine the  form  of  the  bar  so  that  the  horizontal  sections  may  be  pro- 
portionate to  the  stress/  to  which  they- are  subjected.  {Note. — Such  a 
bar  is  a  bar  of  uniform  strength.) 

Ans.  Sectional  area  at  distance  .«•  from  origin  =  — e 

135.  Find  the  upper  and  lower  sectional  areas  of  a  steel  shaft  of  uni- 
form strength,  200  ft.  in  length,  which  will  safely  sustain  its  own  weij^ht 
and  100  tons,  7  tons  per  sq.  in.  being  tlie  woricing  stress. 

Ans.   14.3  sq.  in. ;  17.8  sq.  in. 

136.  A  vertical  elastic  rod  of  natural  length  L  and  of  which  the  mass 
may  be  neglected,  is  fixed  at  its  upper  end  and  carries  a  weight  W\  at 
the  lower  end.  A  weight  Wt  falls  from  a  height  h  upon  W\ .  Find  the 
velocity  and  extension  of  the  rod  at  any  time  /. 

Wi  -f-  lV-\  L      )       \dil 

X  being  measured  from  mean  position  of  {IVi  +  IVi), 

137.  Determine  the  functions  F  kind/  in  Art.  24  wlien  Pi  is  zero,  ami 
also  when  the  rod  is  perfectly  free ;  i.e.,  when.  A  =  o  and  A  =  o. 

138.  An  elastic  trapezoidal  lamina  ^y5CA  of  natural  length  /  and 
thickness  unity,  has  its  upper  edge  AB  {2a)  fixed  and  hangs  verti- 
cally. If  a  weight  IV  is  suspended  from  the  lower  edge  CD  {2/A,  show 
th.1t,  neglecting  the  weight  of  the  lamina,  the  consequent  elongation 

=  ^  -fzr     ■■,  log,—.     If  an  adcitional  weight  is  placed  upon   IF  iu\i\ 

then  suddenly  removed,  show  that  the  oscillation  set  up  is  isochronous 


r 


fF/log, 


and  that  the  time  of  a  complete  oscillation  =  n^  — pr — 
Examine  the  case  when  a  =  b. 


f>) 


i 


Ans.  Ext. 


I  IVl 


--" ;  time  of  oscillation  =  itA/  JU- 


139.  J 
w,  find  li 
work  of  e 


Work 

140.  A 
its  base  A 
its  elonga 
=  unity,  a 

141-  A 
its  upper  e 
striking  th 
(middle  C 
of  the  rod 

142.  Di 
of  6  ft.  radi 
second.  E 
measured  a 

143-  A  ( 
second  abo 
normal  to  t 


144-  In  i 
of  water  of 
denly  check 


E  being  the 
coefficient  o 
pipe  circumf 

145-  A  he 
a  horizontal 
vertical  dept 

If  a  unif( 
position. 

Also  find 


i 


EXAMPLES. 


23' 


139.  If  the  specific  weight  of  the  lamina  in  the  preceding  question  is 

w,  find  how  mucli   it  will  stretch  under  its  own  weij^ht,  and  also  the 

work  of  extension.     Determine  the  result  when  a  =  b. 

I      'uib'T  b       wra+b    wP 

Alls.     —- :—  log  -  +  --7    - — -,  ;   -TT- 

2t  {a  —  bf     ^  a      i,E  a  —  b     zE 


Work  = 


■wP 


\E{a  —  by 


\-"'' 


b"      2 


b^)-b^W^,~ 


140.  An  elastic  lamina  in  the  form  of  an  isosceles  triangle  ABC  has 

its  base  AB  (=  la)  fixed  and  hangs  vertically.     If  its  weight  is  W,  fmd 

its  elongation.     Take  coefficient  of  elasticity  =  is,  thickness  of  lamina 

=  unity,  and  L  the  distance  of  C  from  AB.  ^        \VL 

•'  Ans.  . 

6,ah 


141.  A  metal  rod  \  sq.  in.  in  area  and  5  ft.  long  hangs  vertically  with 
its  upper  end  fixed  and  carries  a  weight  of  18  lbs.  at  the  lower  end.  On 
striking  the  rod  it  emitted  a  musical  note  of  264  vibrations  per  second 
(middle  C  of  piano-forte).  Find  the  coefficient  of  elasticity,  the  weight 
of  the  rod  being  neglected.  Ans.  30,979,160  lbs. 

142.  Diameter  of  a  pipe  is  18  in.  ;  at  one  point  it  is  curved  to  an  arc 
of  6  ft.  radius.  Water  flows  round  the  curve  with  a  velocity  of  6  ft.  per 
second.  Determine  the  centrifugal  force  per  foot  of  length  of  elbow 
measured  along  the  axis.  Ans.  124.3  'bs. 

143.  A  disk  of  weight  fFand  area  A  sq.  ft.  makes  n  revolutions  per 
second  about  an  axis  through  its  centre,  inclined  at  an  angle  Q  to  the 
normal  to  the  plane  of  the  disk.     Find  the  centrifugal  couple. 

WAii" 


Ans. 


5.12 


tan  0  ft. -lbs. 


144.  In  a  circular  pipe  of  internal  radius  r  and  thickness  t,  a  column 
of  water  of  length  I,  flowing  with  a  velocity  due  to  the  head  //,  is  sud- 
denly checked.    Show  that 


gh^ 


EtX^ 


|'  +  2K'+l)+$^[' 


E  being  the  coefficient  of  elasticity  of  the  material  of  the  pipe,  E^  the 
coefficient  of  compressibility  of  the  water,  and  /I  the  extension  of  the 
pipe  circumference  corresponding  to  E. 

145.  A  heavy  ball  attached  by  a  string  to  a  fixed  point  O  revolves  in 
a  horizontal  circle  with  a  given  uniform  angular  velocity  gj.  Find  the 
vertical  depth  of  the  centre  of  the  ball  below  the  point  of  attachment. 

If  a  uniform  rod  be  substituted  for  the  ball  and  string,  find  its 
position. 

Also  find  the  position  when  the  ball  is  attached  to  the  fixed  point  by 


'  •  Mum 

■     1i  ' 
1 


f  u 


•lt::;il 


rw% 


1 

1 


232 


THEORY  OF  STRUCTURES. 


a  uniform  rod  ;  r  being  the  ratio  of  the  weight  of  the  rod  to  the  weight 
of  the  ball. 


Ans.  h  = 


=  .C.  .  =  3£ 


GJ- 


//  =  ^^-;/.=^ 


H 

I   +  - 
g  2 


2  W' 


O) 


'         n 
I  +  - 
3 


Ans. 


146.  The  deflection  of  a  truss  of  /ft.  span  is  /  x  .001  under  a  station- 
ar\'  load  W.  What  will  be  the  increased  pressure  due  to  centrifugal  force 
when  If  crosses  the  bridge  at  the  rate  of  60  miles  an  hour? 

242  W 
725  "7"' 

147.  A  fly-wheel  20  ft.  in  diameter  revolves  at  30  revolutions  per 
minute.  Assuming  weigiit  of  iron  450  lbs.  per  cu.  ft.,  find  the  intensity 
of  the  stress  on  the  transverse  section  of  the  rim,  assuming  it  unaffected 
by  the  arms.  Ans.  96  lbs.  per  sq.  in. 

148.  Assuming  15,000  lbs.  per  sq.  in.  as  the  tensile  strength  of  cast- 
iron,  and  taking  5  as  a  factor  of  safety,  find  the  maximum  working  speed 
and  the  bursting  speed  for  a  cast-iron  fly-wheel  of  20  ft.  mean  diameter 
and  weighing  24,000  lbs.,  the  section  of  the  rim  being  160  sq.  in. 

149.  A  60-in.  driving-wheel  weighs  3^  tons,  and  its  C.  of  G.  is  i  in. 
out  of  centre.     Find  the  greatest  and  the  least  pressure  on  the  rails. 

150.  A  wheel  of  weight  W,  radius  of  gyration  k,  and  making  n 
revolutions  per  second  on  an  axle  of  radius  /?,  comes  to  rest  after  having 
made  N  revolutions.     Find  the  coefficient  of  friction.  ■ 

Ans.  sm  0  =  — - — ,  and  coefi.  of  fric.  =  tan  <t>. 

Ng 

151.  A  train  starts  from  a  station  at  A  and  runs  on  a  level  to  a  station 
at  B,  /ft.  away.  If  the  speed  is  not  to  exceed  v  ft.  per  sec,  show  that 
the  time  between  the  two  stations  is 

l_       W_v_  P  +  B 

v"^  g  2  {P-R){B  +  R)' 

W  being  the  gross  weight  of  the  train,  P  the  mean  uniform  pull  exerted 
by  the  engine,  R  the  road  resistance,  and  B  the  retarding  effect  of  the 
brakes. 

Also,  if  the  speed  ['•  not  limited,  show  that  the  least  time  in  which 
the  train  can  run  between  the  specified  points  is 


V 


2/ 


IV 


P  +  B 


g  iP-R)(B  +  R) 
and  that  the  maximum  speed  attained  is 


-  sec, 


-R)(B  +  R) 
P  +  B 


ft.  per  sec. 


152.  A 
a  24-in.  si 
between  t 
locomotiv 
lbs.);  the 
the  wheel; 
friction  be 
distance  i 
attained ;  1 
wheels. 

A 

153.  If 
find  (a)  th< 
is  brought 

'54.  If 
lbs.  per  sq 
hauled  bet 
lbs.  per  toi 
the  train  u 
Also  fii 
drivers,  th 
attained. 

Ans.- 

155-  Th 
locomotive 
diameter  o 
quired  to  s 

156.  Tw 
lbs.),  run  b 
an  average 
the  other  s 

fuel  in  the 

per  hour. 

157.  If  1 
the  wind,  w 
gale  will  dr 
ton. 

158.  A  1 
ried  by  the  ( 
ton,  what  n^ 


%£ 


EXAMPLES. 


233 


152.  A  locomotive  capable  of  exerting  a  uniform  pull  of  2  tons,  with 
a  24-in.  stroke,  20-in.  cylinder,  and  60-in.  driving-wheels,  hauls  a  train 
between  two  stations  3  miles  apart.  The  gross  weight  of  the  train  and 
locomotive  =  200  tons;  the  road  resistance  =  12  lbs.  per  ton  (of  2000 
lbs.);  the  brakes,  when  applied,  press  with  two  thirds  of  the  weight  on 
the  wheels  of  the  engine  and  brake-van,  viz.,  90  tons,  the  coefficient  of 
friction  being  .18.  Find  (a)  the  least  time  between  the  stations  ;  {b)  the 
distance  in  which  the  train  is  brought  to  rest;  (c)  the  maximum  speed 
attained  ;  (rf)  the  pressure  of  the  steam  ;  (e)  the  weight  upon  the  driving- 
wheels. 

Ans. — {a)  513.8  sec.  ;  (b)  990  ft.;  {c)  42  miles  per  hour;  {d)  25 
lbs.  per  sq.  in. ;  (e)  \\\  tons. 

153.  If  the  speed  in  the  last  question  is  limited  to  30  miles  an  hour, 
find  {a)  the  time  between  the  stations;  (<J)the  distance  in  which  the  train 
is  brought  to  rest;  ic)  the  distance  traversed  at  30  miles  an. hour. 

Ans— (a)  543i  sec;  {b)  504^  ft. ;  {c)  7773J  ft. 

154.  If  the  steam-pressure  in  the  above  locomotive  is  increased  to  50 
lbs.  per  sq.  in.,  find  (a)  the  weight  of  the  heaviest  train  which  can  be 
hauled  between  the  stations  in  10  minutes,  the  road-resistance  being  20 
lbs.  per  ton  (of  2000  lbs.)  and  the  braking  power  being  sufficient  to  bring 
the  train  to  rest  in  a  distance  of  720  ft. 

Also  find  {p)  the  braking  power;  (c)  the  weight  thrown  upon  the 
drivers,  the  coefficient  of  friction  being  \;  {d)  the  maximum  speed 
attained. 

Ans. — {a)  3ioitons;  {b)  15.6  tons;  {c)  24  tons;  (rf)  36  miles  per  hour. 

155.  The  weight  upon  the  driving-wheels  {D  in.  in  diameter)  of  a 
locomotive  is  W  tons;  the  adhesion  =  one  fifth  ;  the  cylinders  have  a 
diameter  of  d  in.  and  a  stroke  of  /  in.     Find  the  steam-pressure  re- 


quired to  skid  the  wheels. 


Ans.  400   ,„,   lbs.  per  sq.  m. 
d  I 


156.  Two  trains,  each  with  a  brake-power  of  190  lbs.  per  ton  (of  2000 

lbs.),  run  between  Montreal  and  Toronto,  a  distance  of  333  miles,  against 

an  average  resistance  of  10  lbs.  per  ton.     One  train  runs  through,  and 

the  other  stops  at  N  intermediate  stations.     Show  thai  the  saving  of 

9A^ 
fuel  in  the  former  is  —  per  cent ;  the  speed  is  not  to  exceed  30  miles 

per  hour. 

157.  If  the  end  of  a  railway  wagon  exposes  a  surface  of  6  x  4  ft.  to 
the  wind,  what  is  the  greatest  gradient  up  which  a  20  lb.  to  the  sq.  ft. 
gale  will  drive  it?  Take  the  weight  at  10  tons,  the  friction  10  lbs.  per 
ton.  Ans.   i  in  59. 

1 58.  A  locomotive  and  tender  weigh  70  tons,  of  which  26  tons  are  car- 
ried by  the  driving-wheels.  Taking  the  adhesion  at  \,  friction  10  lbs.  per 
ton,  what  maximum  gradient  can  the  engine  ascend  .'        Am.  i  in  16. 


! 


234 


THEORY  OF  STRUCTURES. 


1 59.  Given  a  locomotive  witii  two  18"  x  26"  cylinders,  the  connectini,'- 

rod  =  6  ft.,  the  boiler-pressure  =  140  lbs.,  and  driving-wheels  of  7'  o" 

,     ,         ,         .1      •       ,  •     .        .         .  .     force  at  ueripliery 

diameter,  calculate  the  adhesion-fnction,  i.e.,  the  ratio  — --, 7--.—  -  . 

weight  on  drivers 

160.  A  railway  wagon  weighing  20  tons,  with  two  pairs  of  wheels 
8'  o"  centre  to  centre,  and  with  its  centre  of  inertia  7'  o"  above  top  ol 
rails,  has  its  wheels  skidded  while  running.  Take  /<  =0.15.  Required 
the  total  retarding  force  and  pressure  of  each  wheel. 

Alls.  7.375  ;  12.625,  ^"<i  3  lois  on  rail. 

i6t.  Find  Ux)  the  least  time  in  which  a  locomotive  exerting  a  uniform 
pull  of  P  tons  can  haul  a  train  weighing  W  tons  between  two  stations 
/  ft.  apart  on  an  incline  of  i  in  ///,  the  brake-power  being  li  tons  and  the 
road-resistance  A'  tons. 

Also  lind  (/')  the  time  between  stations  when  the  speed  is  limited  to 
V  ft.  per  sec. 

■  \  \  'v 


Alls.— (a)  \   2 


where  A  =  A'  -l- 


iv' 


t±IL .  (/.)  L  + 


r  +  n 

\P-A){IJ  +  A)' 


in 


162.  A  locomotive  exerting  a  uniform  pull  of  4  tons  hauls  a  train  of 
200  tons  up  an  incline  of  i  in  200,  between  two  stations  2  miles  apart, 
the  greatest  allowable  speed  being  30  miles  an  hour.  If  the  road-resist- 
ance is  10  Ihs.  per  ton  (of  2000  lbs.),  and  if  the  brakes  are  capable  of  ex- 
erting a  pressure  of  100  tons,  the  adhesion  being  one  fifth,  find  (^)  the 
time  between  the  stations;  (b)  the  distance  in  which  the  train  is  brought 
to  rest ;  (c)  the  distance  traversed  at  30  miles. 

Also,  if  the  speed  is  not  limited  to  30  miles,  find  {d)  the  least  time  in 
which  the  distance  can  be  accomplished  ;  (r)  the  maxirium  speed  attained ; 
(/)  the  distance  in  which  the  train  is  brought  to  rest. 

Alts.— (a)  Si  min.;    (/')  275  ft.;    (c)  7260  ft.;   (d)  4.87  min.; 
(f)  53.8  miles  per  hour;  (/)  880  ft. 

163.  With  the  same  brake-power,  adhesion,  and  road-resistance,  find 
the  weight  of  the  heaviest  train  which  the  locomotive  in  the  preceding 
question,  exerting  the  uniform  pull  of  4  tons,  can  haul  between  the  two 
stations  in  6  minutes.  Ans.  360  tons. 

,  164.  If  the  locomotive  has  60-in.  drivers  and  24-in.  x  20-in.  diameter 
cylinders,  find  the  weight  required  upon  the  drivers  when  the  steam- 
pressure  is  50  lbs.  per  sq.  in.  Ans,  20  tons. 


CHAPTER    IV. 

STRESSES,  STRAINS,  EARTHWORK  AND  RETAINING- 

VVALLS. 


'III' 


f    •  'i 


IHi 


1.  Internal  Stresses. — The  application  of  external  forces 
to  a  material  body  will  strain  or  deform  it,  and  the  particles 
of  the  body  will  be  in  a  state  of  mutual  stress. 

In  the  following  calculations  it  is  assumed  : 

(a)  That  the  stresses  under  consideration  are  parallel  to  one 
and  the  same  plane,  viz.,  the  plane  of  the  paper. 

{d)  That  the  stresses  normal  to  this  plane  are  constant  in 
direction  and  magnitude. 

(c)  That  the  thickness  of  the  plane  is  unity. 

Dc/.  The  angle  between  the  direction  of  a  given  stress  and 
the  normal  to  the  plane  on  which  it  acts  is  called  the  obliquity 
of  the  stress. 

2.  Simple  Strain. — The  solid  ABCD  (Fig.  207)  of  uniform 
transverse  section  A  is  acted  upon  in  the  direction  of  its  length 
by  a  force  P  uniformly  distributed  over  its  end, 

p 
producing  an  intensity  of  stress  -   =/.     At  any 

/I 

other  transverse  section  inn  the  intensity  must  be 
the  same  in  order  that  equilibrium  may  be  main- 
tained. 

Draw  an  oblique  plane  m'n',  inclined  at  an 
angle  0  to  the  axis.  The  total  stress  on  m'n'  =  P 
and  necessarily  acts  in  the  direction  of  the  axis. 

P 
The    intensity  of    the   stress  on  inn' 


m'n' 


P  P 

3  =  — r  sin  6*  =  />  sin  B. 

inn  cosec  u      A 


The  normal  com- 


ponent  of  the  intensity  on  m'n'  =  p  sin'  6  =  pj. 


Pig.  207,. 


235 


f'J 


*1 


236  THEORY  OF  STRUCTURES. 

The  tangential  component  or  shear  on  m'n' 
=  /  sin  6*  cos  0  =  //. 

So,  if  m"n"  is  an  oblique  plane  perpendicular  to  m'n',  the  nor- 
mal component  of  the  intensity  on  7n''n"  =  p  cos"  B  =  p„". 
The  tangential  component  or  shear  on  m"n" 

=zp  cos  6  sin  6  =  p/'. 
...  p^'  _|_  pj'  =  p     and  p/  =  pr  =ps\ne  cos  d  =  P_^}^, 

The    shear  is  evidently   a  maximum   when    2B  =  90°   or 

e  ^  45°. 

3.  Compound  Strain. — {a)   First  consider  an  indefinitely 
small  rectangular  element  OACB  (Fig.  208)  of  a  strained  body, 
p       p       p      p  ^^Pt   "'   equilibrium    by   stresses 

^^      \      \      \  acting  as  in  the  figure. 

'         ■       '"        '    ip  p  is  the  intensity  of  stress  on 

-"■''^  the  faces  OB,  A  C,  and  a  its  ob- 
-O'/  liquity. 

"'         q  is  the  intensity  of  stress  on 
'^     '"^         the  faces  OA,  EC,  and  /?  its  ob- 
FiG.  208.  liquity. 

OB  .p  cos  a,  the  total  normal  stress  on  OB,  is  balanced  by 
AC.p  cos  a,  the  total  normal  stress  on  AC. 

OB  .p  sin  a,  the  total  shear  on  OB,  is  equal   in  magnitude 
but  opposite  ii   direction  to  AC .p  sin  a,  the  total  shear  on  AC. 
These   two    forces,  therefore,   form   a  couple  of   moment 
OB  .p  sin  a.OA. 

Similarly,  the  total  normal  stresses  on  the  faces  OA, 
BC  balance  and  the  total  shears  form  a  couple  of  moment 
OA.  g  sin /3.  OB. 

In  order  that  equilibrium  may  be  maintained  the  two 
couples  must  balance. 


or 


.-.  OB  .p  sin  a.OA  =  OA.q  sin  /?.  OB, 

p  sin  a  T=  q  sin  ft  =  t,  suppose. 


COMPOUND   STRAIN. 


^17 


B 

Fig.  209. 


Hence,  at  any  point  of  a  strained  body,  the  intensities  of  the 
shears  on  any  two  planes  at  right  angles  to  each  other  are  equal. 

(d)  Next  consider  an  indefinitely  small  triangular  element 
OAB  (Fig.  209)  of  the  strained  body,  bounded 
by  a  plane    AB  and  two  planes  OA,  OB  at 
right  angles  to  each  other. 

Let  p  be  the  intensity  of  stress  on  OB, 
a  its  obliquity. 

Let  g  be  the  intensity  of  stress  on  OA, 
li  its  obliquity. 

Let  t  be  the  intensity  of  shear  on  each  of 
the  planes  OA,  OB.    Then 

^  =  /  sin  «  =  ^  sin  ft. 

p„  ,  the  normal  component  of/,  =  /  cos  a. 
q„ ,    "         "  "  "  ^.  =  ?  cos  /3. 

Produce  OA  and  take  OC  =  p„  .  OB  -}-  t .  OA  =  the  total 
force  on  OB  in  the  direction  of  OA. 

Produce  OB,  and  take  OD  =  q„.  OA -\- t .  OB  =  the  total 
force  on  OA  in  the  direction  of  OB. 

Complete  the  rectangle  CD. 

OE  represents  in  direction  and  magnitude  the  resultant  of 
the  two  forces  OC,  OD,  and  must  therefore  be  equal  in  magni- 
tude and  opposite  in  direction  to  the  total  stress  on  AB. 

Let/,,  be  the  intensity  of  stress  on  AB.    Then 


(/, .  AB^'  -  OE' 


OC  +  OD'  =  {p„.OB  +  t.  OA)" 
+  (<?„  .OA  +  t.  OBy  ; 


Pr 


OA.OB^       ,       ^   ,     , 
2t p^^-(/„ -f- 5,„)  4  /-. 


^''   \AB)     '   ''"  \AB1     '   ""      AB' 
Let      be  the  angle  between  AB  and  OA.     Then, 


WKL 

^^ 

HH 

■'■TiT) 

'  <  1 

[ 
,   1 1 

'■     <       '    <  ^ 
! 

,•1' 


-ii 


1 


il 

i! 

•  '-    ■  i-. 

;     i       ■ 

Pr  —  Pn   sin'  /  +  q^  cos"      +  2/  sin  y  cos  y  {pn  +  ?»)  +  ^'• 

This  gives  the  intensity  of  stress  on  any  plane  AB  inclined 
at  an  angle  y  to  OA,  and  in  tlie  limit  AB  is  a  plane  through  O. 


•:;r.fH 


WT^.     ' 

m 

Wm 

■ 

•,(t.  JI^^^^^^H 

^38 


THEORY  OF  STRUCTURES. 


J. 


Example. — Consider  an  indefinitely  small  triangular  ele- 
ment abc  (Fig.  210)  of  a  horizontal  beam  bounded  by  a  plane 

b   t.ah's.ai  ~  ^^  inclined  at  0  to  the  vertical,  the 

horizontal  plane  ab,  and  the  ver- 
tical  plane  ac. 

The  element  «^c  is  kept  in  equi- 
librium by  the  stress/  .  ac  upon  ac, 


Subs 


a 


\-^-K 


% 


Fig.  210. 


the  shear  s.ad  (=  t.ab)  along  ad,  the  shear  t .ac  along  ac,  and 
the  stress  developed  in  the  plane  be.  The  weight  of  the  element 
is  neglected  as  being  indefinitely  small  as  compared  with  the 
forces  to  which  it  is  subjected.  Let  the  stress  upon  be  be 
decomposed  into  two  components,  the  one  X .  be  normal  and 
the  other  Y .  be  tangential  to  be. 

Resolving  perpendicular  and  parallel  to  be. 


and 


•or 


.and 


X  .be  —-  p  .ae  cos  0  —  t .  ab  cos  6  —  t .  ac  s\x\  6 
V.  be  =  p.  ac  sin  6  —  t .  ab  sin  0  -{- 1  .ac  cos  0, 

X  =  p  co-J' 6  -  t  sXn  2f) (i) 


Y  =  p 


sin  26 


+  t  cos  26 (2) 


Eq.  ( 
kind  as 

stress  is 

Eq.( 
The] 

article)  ii 

Let  6 
tively  mi 


and 


The  vaiue  of  B  for  which  X  is  a  maximum  is  given  by 

rfV  of 

j    —  o  =  —  p  sin  20  —  2i  cos  26,     or     tan  26  =  —  —.     (3) 
Substituting  the  value  of  B  in  eq.  i,  ve  have 

P  // 

max.  value  of  ^  =  —  -|-  a  /  —  -f-  /'.      .      .     (4) 


The  value  ot  0  for  which  Y  is  a  maximum  is  given  by 
—7  =  0  =  /'  cos  2^  ~  2/  sin  20,     or     tan  20  —      . 

<t0  '  -  2t 


(5) 


Henc( 
which  th 
plane  upc 
ninm,  is  c 

Again 

^  =--  45°. 

Thus, 
axis  at  an 
at  0°,  wl 
the  skin  s 

Fig.  2 
intensity, 
ferriiig  re< 


'■•M 


COMPOUND   STRAIN. 
Substituting  the  value  of  B  in  eq.  (2),  we  have 


239 


max.  value  of  Y 


x/^^" 


(6) 


Eq.  (4)  gives  the  maximum  intensity  of  stress  of  the  same 
kind  as  /.     The  maximum  intensity  of  the  opposite  kind  of 


stress  IS  — 


\/f 


^f 


Eq.  (6)  gives  the  maximum  intensity  of  shear. 

The  position  of  the  planes  of  principal  stress  (see  following 

2t 

article)  is  given  by  tan  2^=  — . 

/ 
Let  ^, ,  ^j  be  the  values  of  B  for  which  .A' and  Fare  respec- 
tively maxima.     Then 


tan  26^,  tan  2^',  = 


2t  p 
p  2t 


=  I, 


and 


.-.  ^-B^  =  45°. 


Hence,  at  any  point,  the  angle  between  the  plane  upon 
which  the  normal  intensity  of  stress  is  a  maximum  and  the 
plane  upon  which  the  tangential  intensity  of  stress  is  a  maxi- 
mum, is  equal  to  45°. 

Again,  /  Is  zero  when  d^  —  90°  or  0°,  and  p  is  zero    vnen 

K  =--  45°. 

Thus,  the  curve  of  greatest  normalintensity  cuts  the  neutral 

axis  at  an  angle  of  45°,  one  skin  surface  at  90°  and  the  opposite 
at  0°,  while  the  curve  of  greatest  tangential  intensity  cuts 
the  skin  surfaces  at  45°,  and  touches  the  neutral  axis. 

Fig.  211  serves  to  illustrate  the  curves  of  greatest  normal 
intensity.  There  are  evidently  two  sets  of  these  curves,  re- 
ferring respectively  to  direct  thrust  and  direct  tension. 


\       I  ('    iuf'^ 


.. 


240 


THEORY  OF  STRUCTURES. 


Fig.  212    illustrates   the    curves  of   greatest  tangential  in- 
tensity. 


90? 


'dti 


JccV    '    ihn^c^^yi 


\   1^1         MS 


-?^v 


s    / 
/  \ 


jl — — ■ 


^^. 


Xet |_  terig'lJ^''^    ,w: 


'^^''"-l.-^K 


:v--^ 


Fig.  211 


Fig.  212. 


4.  Principal  Stresses. — Suppose  that   there  is  no  shear 
on  AB,  Fig.  209,  and  that  the  stress  is  wholly  normal. 
In  such  a  case  OE  must  be  perpendicular  to  AB. 


tan  y  =  cot  COE  = 


qc 

CE 


OC 
OD 


p„.OB-\-t.  OA 


q„  .OA-\-t.  OB 
p,.  tan  y  -\-  t 
q„  +  t  tan  / 


{; 


2/ 


2  tan  y 
q„  —  A~  I  —  tan'  y 


=  tan  2y. 


(7) 


-lE         Two  values  of  y  satisfy  this  equation,  viz., 
y  and  y  -f-  90°. 

Hence,  at  any  point  of  a  strained  body,  there 
are  two  planes  at  right  angles  to  each  other,  on 
whicfi  the  stress  is  wholly  normal. 

Such   planes  are    called   planes   of  principal 
stress,   and    the    stresses    themselves   principal 
stresses. 
^  ^         5.  Ellipse  of  Stress. — At  any  point  of  the 
Fig.  213.  strained  body,  consider  a  small  triangular  ele- 

ment OAB  (Fig.  213),  OA  and  OB  being  the  planes  of  principal 
stress, 

Let/,  be  the  principal  stress  normal  to  OB. 

II     A    11     <t  ((  (I  <(         i<   f^/i 


'is 
If 


CONSTANT  COMPONENTS  OF Pr 


241 


Complete  the  construction  as  before,  and  let  ^  be  the  angle 
between  OE  and  OC.     Then 


.     ,       CE      OD      p,OA      p. 


OC 
COS  ^  =  ^ 


P.OB     A    . 

a:^^^a""^^= 


(8) 


A  sin  rb       .  A  cos  tb 

.'.  cos  y  =  — ,    sin  y  =  — ;  and 

A  Pi 

T  _  c;«»  ,,  _L  .^.^          (^'-  ^°^  '^'^'  _l(^'-  ^'"  ^)' 
I  =  sin   y  -\-  cos   y  = ri +  


A" 


/, 


(9) 


Take  OR  to  represent  A  •"  direction  and  magnitude. 

Let  X,  V  be  the  co-ordinates  of  R  with  respect  to  O.    Then 

X  =  Pr  cos  ?/',     Y  —  py  sin  ^, 

and  eq.  (9)  becomes 

X\Y^  .    ^ 

I  =  —  4-^^ (10) 


XI      Y^ 

p:  ^  p:  ' 


the  equation  to  an  ellipse  with  its  centre  at  O,  and  its  axes 
(equal  to  2A  and  2A)  lying  in  the  planes  of  principal  stress. 
This  ellipse  is  called  the  ellipse  of  stress,  and  the  stress  on  any 
plane  AB  at  O  is  the  semi-diameter  of  the  ellipse  drawn  in  a  di- 
rection making  an  angle  ^p  with  the  axis  OC,  tp  being  i     en  by 

i) 
td.\\ip  =  —  co\.  y.     (Eq.  (8).)    .     .     .     .     (ll) 
Pi 

6.  Constant  Components  of  Pr.  —  Take  the  planes  of 
principal  stress  as  planes  of  reference  (Fig.  214). 


•n. 


'.  '^^ 


Fig. 


S«f'f 


Tr 


m 


'I 


Hf 


242  .     THEORY  OF  STRUCTURES. 


Draw  (9A''perpL.idicular  to  AB,  and  take  ON  = 


P.-^P. 


Let  the  obliquity  oi  OR  =  d  =  RON  z=:  go°  —  ij,  —  y, 
"'       Join  NR.    Then 


NR'  =  (?/?'  +  ON'  -  2(?i^ .  ON  cos  ^ 
:      =  A'  +  (— t"^-)'  -  A(/.  +  A)  sin  (^  +  y). 

But    Z^'  =/,'  sin"  y  -\-p^  cos"  ;/,     and 

p.  p 

sin  {ip  -\-y)z=i  sin  ^  cos  y  •\-  cos  V^*  sin  ;/  =  "^  cos'  y  -\-  --  sin'  y. 

Pr  Pr 

(See  eqs,  (8).) 

.-.  NR'  =  A'  sin"  y  +  A'  cos"  r  +  (^— ^')' 

—  (/.  -\-p^){Px  sin'  K  +A  cos"  y) 


.-.  NR  = 


P.- P. 


(12) 


Hence,  the  intensity  of  stress  OR  at  any  point  0  0/  the  plane 
AOB  is  the  resultant  of  two  constant  intensities 


^31^1^ 


^^^A±A    ^„^^    ^^^h^zP,^ 


the  former  beint;^  perpendicular  to  the  plane. 


THE  ANGLE  ONR. 

7.  The  Angle  onb.  =  2y. 

sin  ONR       OR  p. 


243 


sin  tj  NR       p,  —  /,' 


But 


sin  B  =  cos  (^  4"  K)  =  cos  '/>  cos  y  —  sin  '/»  sin  y 

Px—Pi.  Px—  P-i 

—  sin  ;^  cos  y  = 


/. 


sin  ONR 


2pr 


sih  2;/. 


/,- 


2/,  -^  2 


.-.  sin  6>^V^  =  sin  2;/,     or     0NR  —  2y.     .     .     (13) 

Let  A''A'  (Fig-  214)  produced  in  both  directions  meet  OA  in 
F  and  OB  in  G.  , 

The  angle     OFN  =  180°  -  ONR  -  NOF 

=  180°  —  2y  -  (90°  -y)  =  go°  —  y  =  FON. 

.-.  NF  =  NO;    so,    NG  =  NO  =  NF.   ■ 


Also 


and 


.•.  A'' is  the  middle  point  of  FG. 


RF=  FN  -NR=ON-  NR=p, 


RG  =  RN-\-  NG  =  AW+  ON  =  p,. 


i  ! 


N.  v5.— The  shear  at  O 


P.- P. 


cos  {2y  —  90")  =  (/,  —  p^  sin  ;/  cos  ^. 


..^:n 


THEORY  OF  STRUCTURES. 


8.  Maximum  Shear. — ON  has  no  component  along  AB. 
Hence,  the  shear  on  AB  \s  NR  cos  (angle  between  A^R  and 
AB)y  and  is  evidently  a  maximum  when  the  angle  is  nil.     Its 

value  is  then  NR,  or  ^'"J"^'. 

9.  Application  to  Shafting. — At  any  point  in  a  plane  sec- 
tion of  a  strained  solid,  let  r  be  the  intensity  of  stress,  and  0 
its  obliquity. 

At  the  same  point  in  a  second  plane  let  s  be  tlie  intensity 
of  stress,  and  B'  its  obliquity. 

By  Art.  6,  r  and  s  are  the  resultants  of  two  constant 
stresses 


and 


A±A    ^„j    Aj-A, 


...  {^)'=  f-4-^-=)'+  .-  -  KA  +  A)  cos  e      (,4) 


(^-^f  =  (^&)'+  .■  -  .(A  +  A)  cos  »'.     (.5) 


and 


Subtracting  one  equation  from  the  other, 

v  =A  +A 


r'  —  r 


r  cos  6^  —  5  COS  0 


.     .     .     (16) 


First.  Consider  the  case  of  combined  torsion  and  bending, 
as  when  a  length  of  shafting  bears  a  heavy  pulley  at  some  point 
.„    between  the  bearings. 

Let  p  be  the  intensity  of  stress  (compression  or  tension) 
due  to  the  bending  moment  J/4. 


■  ft 

APPLICATION   TO   SHAFTING. 

245 

1 

Let  q  be  the  intensity  of  shear  due  to  the  twisting 

nno- 

ment  Mt . 

p  and  q  act  in  planes  at  right  angles  to  each  other. 

( 

.-.  r  cos  Q  —  pt 

rs\nd  —  q-s,    ar  .'     6'  —  90''. 

4 

.'•r'=f-\-Q- 

and    s  =  q. 

Hence,  by  eq.  (i6), 

\ 

P.^-P.^P\ 

(17) 

and  by  eq.  (15), 


{P^P-^U+f; (,8) 


••A  =  5+Vt+^ 


(19) 


and 


=  i-\/f+^ (^) 


The  max.  shear  = 


A -A 


=/ 


+  ?';  ....   (21) 


also 


p=^  (Chap. VI.)    and    ?  =  ^'      (Chap.  IX.) 


for  a  shaft  of  radius  r. 


-'-P.  =  ^AM,-^  V~M,^  +  Mt\',     •    '    .    (22) 


ll'fi'       !   »   f  -it 


w.!>aWI£ 


246 
and 


THEORY  OF  STRUCTURES. 


A  -P 


2  Ttr  *  n~      / 


(2.3> 


M 


--LA 


Perhaps  the  irtost  important  example  of  the  application  of 

the    above    principle  is  the  case  of  a 

shaft  acted  upon  by  a  crank  (Fig.  2 1 5). 

A  force  /'applied  to  the  centre  6"  of 

the   crank-pin   is  resisted   by  an   equal 

and  opposite  force  at  the  bearing   B, 

forming  a  couple  of  moment  P.  CB  =  M. 

This  couple  may  be  resolved  into  a 

F'G.  21S.  bending  couple  of  moment  Mt  =  P.  AB 

=  P .  EC  cos  S  =  M  cos  d,  and  a  twisting  couple  of  moment 

Mt  =  P  .  AC  =  P  .  BC  sin  6  =  M  sin  6  ;    d  being  the  angle 

ABC. 

■   .:p,  =  -^\Mcos6-\-M\  =  ^cos^^;  .    .    (24) 


^j^^ 


Ttr 


nr 


and  the  max.  shear  = 


2M 


7ir 


.     .     .     (25) 


If  the  working  tensile  or  compressive  stress  (/,)  and  the 
working  sKear  stress   [— -j  are  given,  the  corresponding 

values  of  r  may  be  obtained  from  eqs.  (22)  and  (23)  or  eqs. 
(24)  and  (25) ;  the  greater  value  being  adopted  for  the  radius 
of  the  shaft. 

Second.  Consider  the  case  of  combined  torsion  and  tension 
or  compression. 

Let  the  tensile  or  compressive  force  be  P. 

A  the  intensity  of  the  tension  or  compression,  =  —5 ; 


shear 


2M^ 

Ttr' 


S'Jj-i 


PRINCIPAL  AND   CONJUGATE   STRESSES. 


247 


and 


•••A=^.j-V-+^'!-  •  •  <^^^ 


li  t* 


P. -p. 


2itr    y  '      r'  ^     ' 


10.  Conjugate  Stresses. — Consider  the  equilibrium  of  an 
indefinitely     small     parallelopiped 
abed  {¥\g.  216)  of  a  strained   body, 
the  faces  ab,  cd  being  parallel  to 

the  plane  XOX,  and  the  faces  ad,    Y — -^ -0'- 

hc  to  the  plane  YO  Y. 

Let  the  stresses  on  ab,  cd  act 
parallel  to  the  plane  YOY.  The 
total   stresses    on   ab   and   cd  are  '''°-  "*• 

equal  in  amount,  act  at  the  centres  of  the  faces,  are  parallel  to 
YOY,  and  therefore  neutralize  one  another. 

Hence  the  total  stresses  on  ad  diwd  be  must  also  neutralize 
one  another.  But  they  arc  equal  in  amount,  and  act  at  the 
middle  points  of  ad,  be;  they  must  therefore  be  parallel  to 
XOX.  ,         , 

Hence,  if  two  planes  traverse  a  point  in  a  strained  body, 
and  if  the  stress  on  one  of  the  planes  is  parallel  to  the  other 
plane,  then  the  stress  on  the  latter  is  parallel  to  the  first 
plane. 

Such  planes  are  called  planes  of  conjugate  stress,  and  the 
stresses  themselves  are  called  conjugate  stresses. 

Principal  stresses  are  of  course  conjugate  stresses  as  well. 

Conjugate  stresses  h.uc  equal  obliquities,  each  obliquity 
being  the  complement  of  the  same  angle. 

II.    Relations     between     Principal     and     Conjugate 

Stresses  (Fig.    217).— Take  any  line  C^TV  =^^^-^^  . 


11 

If'''  . 


^1 
■•■  1 '. 


ii 


THEORY  OF  STRUCTURES. 


P  — P 

With  N  as  centre  and  a  radius  =  — ^ ,  describe  a  semi- 

2 

circle.  • 

Let  Q  be  the  common  obliquity  of  a  pair  of  conjugate 
stresses. 


Fig.  ai7. 

Draw  ORS,  making  an  angle  H  with  ON,  and  cutting  the 
semicircle  in  the  points  R  and  S. 

Join  NR,  NS. 

OR  and  OS  are  evidently  a  pair  of  conjugate  stresses. 

Draw  NV  perpendicular  to  RS  and  bisecting  it  in  V. 
I      Draw  the  tangent  OT;  join  NT. 
]      Let  OR  =  r,  0S  =  s.    Then 

rs-OR.OS=  OT*  =  ON'  -  NT*  ' 


=  (n4-M)"=AA,    (.3) 


and 


r-\-  s  =  2OV  =  2ON  cos  ^  =  (A+A)cos  6.      (29) 


The  maximum  value  of  the  obliquity,  i.e.,  of  6,  is  the  angle 
TON. 

Call  this  angle  0.    Then 


NT      p,—p,  ,    , 

sm0  =  7r-.7  =  T-rT (30) 


ON      A+A 


PRINCIPAL  AND   CONJUGATE   STRESSES. 

Let  OR,  OR  be  a  pair  of  con- 
jugate stresses  (Fig.  218). 

Let    OG,   OH  be   the   axes  of   r; 
;4ieatest  and  least  principal  stress, 
respectively. 

Draw  ON  normal  to  OR'. 

Let  the  angle  GOR  =  ^,  RON 
=  e,  HON=  GOR'  =  y,  as  before.    Then 


249 


f  =  gO°-y-e; 


and  by  eqs.  (8), 


~  cot  y  =  tan  ip  =cot  {y-\-ff)', 

.  cot(r  +  ^)  _  A 
cot  X  A' 


sin0= 


sin  6 


or 


A  -A  _  coty-cot(>^  +  g)  _    

A+A       cot  >' +  cot  (;/ -f  6^)        sin  (2;/ 4-^)" 

sin  ^ 

.••  sin  (2y  4-  p)  =  -^ — :r, 
^  '^    '     '      sin  0 


>'  =  i{-*  +  M^)l <3') 


Hence, 


mgle  ^W=  90°  -  y  =  ^{  i8o°+^-sin-(|^)  }    (32) 


and 


anglcffOR=y  +  d    =  1 1  ^  +  sin-(~^)  | .  .    .    (33) 


mi 


r 


\i 


250  THEORY  OF  STRUCTURES. 

12.  Ratio  of  V  to  s. 

l_qR  _  OV-RV  _  OV-VNR'-NV* 
s  ~  OS  ~  OV i-RV  ~  oy-L.  {^NR'  —  NV 

_  ON  cos  a  -  \/NW^^~W^rn) 

~  ON  cos  B  '^^VnW^'WN^^^^ 

NR      A -A      A^T-        .     _..       .     , 
7TT7  =  -     ,        =  TTiTT  =  Sin  TON  =  sin  0. 
ON     A+A      <^^ 

r       cos  0  —  l/sin'  0  —  sin"  6 
"  s  ~  cDse+  i/sin'^  0  -  sin'  ^ 


But 


cos 


COS 


cos  ^  —  i/cos"  6^  —  cos"  0 
COS  ^  +  ^cos'  6  —  cos'  0 


•      •         •         • 


(34) 


Let 


cos  0 
COS  8 


=  sin  Of.    Then 


I  T  cos  a 
I  ±  cos  or 


=  tan 


a  a 

'  -    or     =  cot'  - , 


(35) 


If   ^  =  0,  «   =  90  —    0. 


.-.  J  =  tan"  \4S 


1(0=0,  a  =z  90°. 


(45 -f)     or     =cot'(45-f).   .    (36) 


r  _ 
s  ~ 


(37) 


RELATION  BETWEEN  STRESS  AND   STRAIN. 


251 


13.  Relation  between  Stress  and  Strain. — Let  a  solid 
body  be  strained  uniformly,  i.e.,  in  such  a  manner  that  lines  of 
particles  which  are  parallel  in  the  free  state  remain  parallel  in 
the  strained  state,  their  lengths  being  altered  in  a  given  ratio, 
which  is  practically  very  small.  Lines  of  particles  which  are 
oblique  to  each  other  in  the  free  state  are  generally  inclined  at 
different  angles  in  the  strained  state,  and  their  lengths  are 
altered  in  different  ratios. 

Let  the  straining  of  the  body  convert  a  rectangular  portion 
ABCD  (Fig.  219)  into  the  rectangle  AB'C'D',  where  AB'  = 
(I  +  a)AB  and  AD'  =  (i  +  fi)AD. 

Now  a  and  fi  are  very  small,  so  that  their  joint  effect  may 
be  considered  to  be  equal  to  the  sum  of  their  separate  effects. 
Hence :  •  ' 

First.  Let  a  simple  longitudinal  strain  in  a  direction  paral- 
lel to  AB  convert  the  rectangle  ABCD 

into  the  rectangle  ABED,  where  BB'  ^\ .j'X':..K_.C' 

=  a.AB. 

A  line  OF  will  move  into  the  posi- 
tion OF',  where  FF'  =  a .  DF,  and 

OF'  -  OF 


the  strain  along  OF  = 
FF' cos  d      a.BFcosff 


OF 
=  a  cos'  8, 


"b' 


Fig,  aig. 


""        OF  OF 

0  being  the  angle  OFD. 

Also,  the  "  distortion  or  deviation  from  rect angularity" 


=  angle  FOF'  = 


FF'9,me      a.DFsmO 


OF 


OF 


=  a  cos  ^  sin  ^. 


Seco7id.  Let  a  simple  longitudinal  strain  in  a  direction 
parallel  to  AD  convert  the  rectangle  ABCD  into  the  rectangle 
ABKD',  where  DD'  =  fi.AD. 

The  line  OF  will  move  into  the  portions  O'F",  where 
00'  =  fi.AOdindF"F=DD'  =  ^.AD. 


the  strain  along  OF  = 


O'F' 


OF 


OF 


4  "ife.  '\ 


252  THEORY  OF  STRUCTURES. 

Draw  O'M  parallel  to  OF.    Then 

O'F"--  0F=  O'F"-  0'M=  F"M sin  fi  =  {F"F-FM)  sin  0 
=  {DV  -  00')  sin  e  =  ti{AD  -  AO)  sin  0 
=  ft.  on  sin  tf. 

.      ,        ^ ^      ft.OD  sin  e       ^  .  .  ^ 
.'.  the  strain  along  OF  = ^jx =  /?  sin  0. 


RE 


Consid( 
/,  and  />, 
///;,  BC  ai 
Chap,  ill, 


The  distortion  =  the  anrrlc  F"0'M 


F"I\f  cos  0      fi.ODcos  )       ^    .     , 

; =■  p  sni  t)  cos  c'. 


6>i'^ 


OF 


Hence,  when  the  strains  are  simultaneous,  the  line  OF  yj'xW 
take  the  position  OF  ''  between  O'F" d^nd  OF',  and 

the  total  strain  along  OF  =^  a  cos'  0  -\-  fi  sin'  ^  ; 
the  total  distortion  =.{a  —  fi)  sin  ^  cos  0, 

Again,  draw  a  line  OG  perpendicular  to  OF. 

The  angle  OGA  —  90°  —  (f,  and  hence,  from  the  above, 

the  total  strain  along  OG  —  a  sin'  G  -^  /3  cos^  6, 

and  the  corresponding  distortion  —  {a  —  /?)  sin  6^  cos  6^. 

Denote  the  strain  along  OF  by  <•, ,  that  along  OG  by  ^, ,  and 
each  of  the  equal  distortions  by  t.    Then 

e,  4-  r,  =  «  -f  /y. 

Again,  if  OF,  OG  are  the  sides  of  a  rectangle  enclosed  in 
the  rectangle  A  BCD,  the  straining  will  convert  the  rectangle 
into  an  oblique  figure  with  its  opposite  sides  parallel.  The 
lengths  of  adjacent  sides  are  altered  by  the  amounts  f,  and  e, , 
and  the  angle  6'  by  2t.  The  above  results  may  also  be  consid- 
ered to  hold  true  if  the  straining,  instead  of  being  uniform, 
varies  continuously  from  point  to  point. 


and  the  str 
Uici:  A/iCl 

If  the  s 
i.e.,  if  the  o 

A-- 

.'.  a  = 
Thus  th 


so  that  the  1 

Also,  if 
point  of  AL 


^.  = 


and  the  dist( 


i? 


RELATION  BETWEEN  STRESS  AND   STRAIN.  253 

Consider  a  unit  cube  A  BCD  subject  to  stresses  of  intensity 
/I,   and  /,   upon   the    parallel   faces 
///;,  nC  and  AB,  DC.     By  Art.  3,                       [" 
Chap,  ill,  Di i -C 

P.       A 


a  ■=■ 


E      mE ' 


^  ~  ni/i  ^  E  ' 


>P, 


K 


Fig. 


and  the  strain  perpendicular  to  the 

hicc  A ncjj=  -\-A-. 

ntE      VI  It 

If  the  stresses  are  of  equal  intensity  but  of  opposite  kind^ 
i.e.,  if  the  one  is  a  tension  and  the  other  a  compression, 


/.  =  —  A=/.  suppose. 
E^ 


.'.  a  =  —  /?  —  y.l  1  4-  - ),  and  the  third  strain  is  ni/. 


Thus  the  volume  of  the  strained  solid 

=  (l  4-  «)(!  —  ^)(0  =  I  —  «'  =  I,  approximately, 

so  that  the  volume  is  not  sensibly  changed. 

Also,  if  OGHF  is  an  enclosed  square,  O  being  the  middle 
point  of  AD,  6  =  45°,  and 


e^  =  ej=  =  o  =  strain  along  OF  or  OG, 


and  the  distortion  .—  change  in  angle  O 


2t  =  2 


a  —  p 


=  "  =  |(-+i-)- 


254  7'HKOKV  (>/•■  s'rA'UcruA'/-:.s. 

This  result  may  be  at  once  cli;cliicc<l  from  the  figure.     I'or 


tan 


I^Oh        on        I     \-  fi        \  ~-  a  /no"  -  2A 

=  -»,'/,  =  — ,—  -  =    ■  r  ^  =  tan  K , 

2  hD      I  -f-  tr       \  -\-  a  \      2       / 


or 


I  —  tf       1  —  tan  t      I 


I  -f  'f  ""  I  +  tan  /  ~  1  -f-  /  ' 
since  /  is  very  small,     llcnce 

t  -  n. 

As  already  shown  in  A:'  3,  shearing  cannot  take  plaio 
along  one  pl.me  only,  .iiul  at  any  i)o>nt  of  a  strained  solitj  tin 
shears  along  plan  -s  at  right  angles  are  of  ecpial  intcnsitj'.  Tlu' 
elTect  of  such  stresses  is  merely  to  pioduce  a  (distortion  c/ 
_/fj;'"//;r,  and  generally  without  sensible  change  of  volume. 

Thus,  shears  of  intensity  .v  along  the  parallel  faces  of  t'u' 
unit  square  A  BCD  will  merely  distort  the  square  into  a  rhom 
bus  yI^>C"/^' (Fig.  2j0.      Denoting  tlu-  change  of  angle  by  2/, 
S  and  assuming  that  the  "stress   is  propui- 

■  C'  tional  to  the  strain," 

s  ~  G.2t, 

where  G  is  a  coefficient  calleil  the  modulus 
of  transverse  elasticity,  or  the  eoejfieient  of 
rigidity,  and    ilepeiuls    upon   a  e/iangr  oj 

Fig.  aji.  forill. 

Consider  a  section  along  the  diagonal  BD. 

The  stresses  on  the  faces. /A'.  A  J),  and  on  CB,  CD,  resolved 
parallel  and  perpendicular  to  BD,  are  evidently  equivalent  to 
nil  and  a  normal  force  .v  i  2,  respectively.  Thus,  there  is  no 
sliding  tendency  along  JU),  but  the  two  portions  ABD  and 

s  \  2 
CBD  exert  upon  each  other  a  pull,  or  tension,  of  intensity   -77% 

_  i^  — 
\  2 

Similarly  it  maybe  shown  that  there  is  no  tendency  to  slide 
.along  AC,  but   that   the   two  portions  .IBC  and   ADC  exert 


,, 

1 

1 
1 

1 
1 
1 
1 

1 

/ 

A        ^   -e" 

B 

<ipon  e.icl 
lo  tile  si 
duced  by 
.'I  45".     J 


aiul 


Now  ;/, 

Again, 

(Art.  23),  i 


and  licnce 


14-  Rai 

work  tends 
Ran  kin  I 
<lirections  t 
lo  the  gene 
(2^  that  i\u 
■iiid  that  the 
only. 

In  the  ' 
llie  horizon 
is  sometime: 
Adoptiuj 
librium  of  tl 
pressure  bet 
by  a  plane  s 
less  than  the 


SB3HUJiap«r 


A'.IXA'/A/-/':'.S  EARTinVOKK    TllF.ORY, 


255 


»i[)«)ii  cull  oilier  ii  i)icsHurc  of  iiitiMisily  .v.  The  straininj^'  due 
to  tlie  slicaiinjf  stresses  is,  therefore,  ideiitical  with  that  pro- 
(hieed  by  a  thrust  and  tension  of  va\\\\\\  intensity  upon  planes 
at  45  '.      Hence,  as  provetl  above, 


I'.  \         ml 


aitd 


a 


P.       Ill       _   s 
Z  I     f   VI  ~~  2/' 


Now  ///  raiel)'  exceeds  4,  and  hence  (1  is  ^'cnerally  <  ^/;\ 
Aj^ain,  the  coijjhiciit  of  elasticity  of  voliiiiir,  or  tidiic  elasticity 
(An.  23),  is 


and  hence 


m 


6K  \   2G 
3  A'—  2G' 


14.  Rankine's  Earthwork  Theory.— A  mass  of  earth- 
work tends  to  take  a  definite  slope. 

Rankine  assumes,  (l)  that  the  stresses  exerted  in  different 
directions  throufjh  a  particle  of  a  granular  mass  are  subject 
to  the  general  principles  enunciated  in  the  preceding  articles  ; 
(2\  that  the  cohesion  of  the  particles  is  gradually  destroyed, 
iiid  that  the  stability  of  the  mass  ultimately  depends  on  friction 
only. 

In  the  ''mit,  therefore,  the  face  of  the  mass  is  inclined  to 
the  horizon  at  an  angle  equal  to  the  angle  of  friction,  or,  as  it 
is  sometimes  called,  the  angle  of  repose. 

Adopting  for  the  present  Rankine's  assumjjtions,  the  equi- 
librium of  the  mass  requires  that  the  direction  of  the  mutual 
pressure  between  the  two  parts  into  which  the  mass  is  divided 
by  a  plane  shall  make  an  angle  with  the  normal  to  the  plane 
less  than  the  angle  of  friction. 


I' 


256  THEORY  OF  STRUCTURES. 

Denote  the  angle  of  friction  by  0. 
The  maximum  obliquity  must  be  ^  0. 
By  eq.  (30), 

'  ■  /,  "^  I  -(-  sin  0 


(38) 


Thus,  if  a  pressure  of  intensity  /,  acts  through  a  mass  of 
earthwork,  eq.  (38)  gives  the  least  intensity  of  pressure /j  acting 
in  a  direction  perpendicular  to  that  of  />,  consistent  with  equi- 
librium. 

The  limiting  ratios  of  a  pair  of  conjugate  stresses  in  a  mass 
of  earthwork  may  also  easily  be  determined. 

By  eq-  (34), 


the  ratio  = 


cos  0  —  i/cos"  ti  —  cos''  0 
cos  B  +  |/cos"  d  —  cos'  0 


(39- 


Hence  the  ratio  cannot  exceed 


cos  B  4-  j/cos"  B  —  cos'^  0 

_     __  '  ,  • 

cos  B  —  V'cos"  B  —  cos"  0 


nor  can  it  be  less  than 


cos  B  —  Vcos'  B  —  cos"  0 


cos  B  -f-  v'cos"  B  —  cos'  0 
If  6*  =  o,  the  ratio  becomes 

I  T  sin  0 


I  ±  sm  0 


(40) 


For  example,  let  the  ground-surface  be  horizontal. 
The  pair  of  conjugate  stresses  become  a  vertical  stress  /, 
and  a  horizontal  stress/, . 


"  /,  =^  r  —  sin  0' 


(41) 


or 


as  in  eq. 

Pressu 

tJie  groun 

horizon  at 

Consid 

Let  s  1 

at  Z?. 

Let  r  b 
at /A 

This  coi 

the  ground 

Take  D. 


DE 
DC 


r 

s 


Then  w., 
magnitude  t 
vertical  plan 
'  '  v'olumc  of 

Join  C£. 

The  intei 

Hence,  t 
prism  nC£ 


Again, 
column  CD. 


S    IS 


m 


«PH^^        ■iV^linpMfWf 


ill 


RANKINE'S  EARTIfWORK    THEORY. 


257 


or 


(42) 


/,  >  I  —  sin  0 
/,  =  I  +sin  0'  ' 
as  in  eq.  (38). 

Pressure  agaiftst  a  Vertical  Plane. — Let  ACB  (Fig.  222), 
the  ground-surface  of  a  mass  of  earthwork,  be  inclined  to  the 
horizon  at  an  angle  0. 

Consider  a  particle  at  a  vertical  depth  CD  =  x  below  C. 

Let  s  be  the  vertical  intensity  of  pressure  on  the  particle 
at  Z*. 

Let  r  be  the  conjugate  intensity  of  pressure  on  the  particle 
at /A 

This  conjugate  pressure  acts  in  the  direction  ED  parallel  to 
the  ground-surface,  and  its  obliquity  is  6. 

Take  DE  so  that 


DE 
DC 


r 
s 


cos  6  —  4  cos'  8  —  cos'  0 


cos  0  -\-  Vcos''  0  —  cos'  0 


Then  w .  ED  represents  in  direction  and 
magnitude  the  intensity  of  pressure  on  the 
vertical  plane  DC  at  the  point  D,  w  being  the  weight  of  a  unit 
of  volume  of  the  earthwork. 

Join  CE. 

The  intensity  of  pressure  at   any  other   point   m   is    evi- 
dently w.  inn,  mn  being  drawn  parallel  to  DE.. 

Hence,  the  total  pressure  on   the  plane  /?C  — weight   of 
prism  DCE 

w.DC.DE        ,^       zv.DC'r 
— cos  V  = cos  r/ 

2  2  S 


wx' 


„„  ..cos  0  —  Vcos  6  —  cos  0 

=  cos  n =^~-  —  , 

2  cos  /^  -)-  V  cos"  &  —  cos'  0 


(43) 


Again,  s  is  the  pressure  du>  to  the  weight  of  the  vertical 
column  CD. 


.•.  s  =r  tt.*:r  COS  tf, 


(44) 


"-'KSl; 


258 

and 


THEORY  OF  STRUCTURES. 


r  z=.  wx  cos  ^ 


cos  B  —  Vcos"  B  —  cos"  0 


cos  0  -\-  y  cos'  B  —  cos  0 


(45^ 


By  means  of  this  last  equation  the  total  pressure  on  CD 
may  be  easily  deduced  as  follows  : 

The  pressure  on  an  element  dx  at  a  depth  x 


=1  rdx  =  zvx  cos  B 


cos  B  —  •i/cos"  B  —  cos'  0 


dx. 


cos  B  -\-  ^coi'  B  —  cos"  0 
.*.  total  pressure  =    /  "rdx  =  etc. 

The  total  resultant  pressure  is  parallel  in  direction  to  the 
ground-surface,  and  its  point  of  application  is  evidently  at 
tivo  thirds  of  the  total  depth  CD. 

15.  Earth  Foundations. — Case  I.  Let  the  weight  of  the 
superstructure  be  uniformly  distributed  over  the  base,  and  let 
p^  be  the  intensity  of  the  pressure  produced  by  it. 

If  //,  is  the  maximum  horizontal  intensity  of  pressure  cor- 
responding to/„, 

A_  <  I  +sin  0 
ph  ~  I  —  sin  0 ' 

In  the  natural  ground,  let  p^  be  the  maximum  vertical  in- 
tensity of  pressure  corresponding  to  the  horizontal  intensity 
//,  •    Then 

Ph_  <  I  +  si"  0 
/„  ~  I  —  sin  0 " 
Hence 

A  <  (^  +sin0 
/,  =  \i  —sin  0 

If  X  is  the  depth  of  the  foundation,  and  w  the  weight  of  a 
cubic  foot  of  the  earth, 


p^  =-wx\ 

.   A</i_  +  sin0y 

zi<x  -  \i  -  sui  0/ 


6) 


cubic  foe 


l"lence,  a 


Case  : 

uniformly 
mum  inter 
By  Cas 


In  the  r 
pressure  is 


When  t 
tends  to  rai; 
<jf  the  super 
^^eight  of  tl 
2ontaI  pressi 


r 


ombining  t 


Combining  ( 


4 


(■ 


t  -r':»'p 


iini 


EAPTH  FOUNDATIONS. 


259 


let.  h  -\   ;r  be  the  height  of  the  superstructure,  and  let  a 
cubic  toot  of   ;•  weigh  lv' .     Then 

A  ^  ^\x  +  h). 
i^ence,  a  minimum  value  of  x  is  given  by 


w'{h-\-x)  _  f\  -\'  sin  0y_   I 
^  Vi  ^^in  0/  ~  iP' 


K;;f 


suppose; 


w  —  ^v  k'' ' 


(47) 


Case  II.  Let  the  superstructure  produce  on  the  base  a 
uniformly  varying  pressure  of  maximum  intensity/,  and  mini- 
mum intensity/,. 

By  Case  I, 


A  <  /'  +sin  0y 
ivx  =  \i  —  sin  0/ 


(48) 


In  the  natural  ground  the  minimum  horizontal  intensity  of 
pressure  is 

I  —  sin  0 

''  I  -f-  sin  0 

When  the  foundation-trench  is  excavated,  this  pressure 
tends  to  raise  the  bottom  and  push  in  the  sides.  The  weight 
of  the  superstructure  should  therefore  be  at  least  equal  to  the 
weight  of  the  material  excavated  in  order  to  develop  a  hori- 
zontal pressure  of  an  intensity  equal  to  //, . 

A  <  I  —  s''^  0 
■  ■  A  =  I  +  sin  0  ■ 

Combining  this  with  the  last  equation, 

A>, . 

ivx  — 
\  ombining  (48)  and  (49), 


(49) 


/,</i4-sin0y 
(Pankine's  Civil  Engineering,  Arts.  237,  239.) 


H-i 


260 


THEORY  OF  STRUCTURES. 


16.  Retaining-walls.—  Consider  a  portion  ABMN  of  a 

wall  (Fig.  223). 

Let  W  be  its  weight,  and  let  the  di- 
rection of  W  cut  MN  in  C. 

Let  P  be  the  resultant  of  the  forces 
externally  applied  to  ABNM  d,nd  tend- 
ing to  overthrow  it.  Let  D  be  its 
point  of  application,  and  let  its  di- 
rection meet  that  of  VV  in  E. 

Let  F  be  the  centre  of  pressure  (or 
y^  M  resistance)  at  the  bed  MN. 

Fig.  323.  Let  0  be  the  middle  point  of  MN. 

Let  MN  =  t,  OF  —  qt,  OC  —  rt,  q  and  r  being  each  less 
than  unity. 

Let  x'  and  y' ,  respectively,  be  the  horizontal  and  vertical 
co-ordinates  of  D  with  respect  to  F. 

Let  the    inclination    to  the    horizon  of    MN  =  a,  of  /"'s 
direction  =  /S. 

Conditions   of  Eqiiilibriian. — {a)  The   moment  of   P  with 

respect  to  F  ^  the  moment  of  W^with  respect  to  F,  or 


P{y'  cos  (i  —  x'  sin  /^)  ^  W{qt  =F  rt)  cos  or ;  . 


(51) 


the  upper  or  lower  sign  being  taken  according  as  C  falls  on  the 
left  or  right  of  O. 

In  ordinary  practice  q  varies  from  \  to  f . 

Example. — A  masonry  wall  (Fig.  224) 
of  rectangular  section,  x  ft.  high,  4  ft.  wide, 
weighing  125  lbs.  per  cubic  foot,  is  built 
upon  a  horizontal  base  and  retains  water 
(weighing  62^  lbs.  per  cubic  foot)  on  one 
side  level  with  the  top  of  the  wall. 


Fig.  224. 


^=62i^- 


JF=  125  X  4-*-,    af  =  0,    /?  =  0,     ^  =  4  ft. 


If     q^ 

If     q^ 

{b)  The 

not   exceed 

cru.shing. 

distributed. 

tensity  of  t 

most  compi 

Let  /  b 

total  pressui 

Three  ca 

Case  I. ; 

fi  om  /  at  M 

Take  Mt 

Tile  pre.s 

is   represent( 

MGN. 

The  ordi 
centre  of  gra 
the  centre  of 


Case  II. 


y  =  3+4y,    -^'~3'    '*~°' 


.3     < 


■•  W-^'  =  2000qx, 


Fig 


RE  TA INING-  WALLS. 


261 


or 


x"  ^  192^. 


(52) 


frnFl 


!!i;*i! 


< 


If     q  —  \,     ;r'  ^  48     and     x  ^  6.928  ft. 
If     ^  =  f,     a'  <  72     and     X  <  8.485  ft. 

(^)  The  inaxiinum  intensity  of  pressure  at  the  bed  MN  must 
not  exceed  the  safe  working  resistance  of  the  material  to 
crushing.  The  load  upon  the  bed  is  rarely  if  ever  uniformly 
distributed.  It  is  practically  sufificient  to  assume  that  the  in- 
tensity of  the  pressure  diminishes  at  a  uniform  rate  from  the 
most  compressed  edge  inwards. 

Let  /  be  the  maximum  intensity  of  pressure,  and  R  the 
total  pressure  on  the  bed. 

Three  cases  may  be  considered. 

Case  I.  Let  the  intensity  of  the  pressure  diminish  uniformly 
from /at  M  to  o  at  A^(Fig.  225). 

Take  MG  perpendicular  to  J/iV  and  =/;  join  GN. 

The  pressure  upon  the  bed    G' 
is  represented  by  the  triangle 
MGN. 


M  F         O  N 

The    ordinate    through    the  Fig.  325. 

centre  of  gravity  of  the  triangle,  parallel  to  GM,  cuts  MN  in 
the  centre  of  pressure  F. 


.'.  qt=^OF=OM-  FM  = 


i  _  1 
3    "  6^' 


GS. 


Case  II.   Let  the  maximum  intensity />  MG  in  Case  I. 
Hn^  Take  MH  —  f,  and  the  triangle 

X.  MHK  =  R  (Fig.  226). 

The  pressure  on  the  bed  is 
now  represented  by  the  triangle 
MHK. 


'V. 


.M 


O 

Fig.  jafi. 


K 


N 


R  =  \MH.  MK=  \f.  MK. 
The    ordinate    through    the 


262 


THEORY  OF  STRUCTURES. 


centre  of  gravity  01  the  triangle  MHK  parallel  to  HM  cuts  MN 
in  the  centre  of  pressure  F. 


qt  =  0F=  OM-  MF=z--  — . 

2  3 


But  MK=Y\ 


t       1  R 

•*•  ^^  =  T 7 ;    and  hence 

2       3  / 

I       2i?       ,  .        .^      ,  I 

(J  —  - y  and  IS  evidently  > 


6' 


(53) 


Case  III.  Let  the  maximum  intensity/  <  MG  in  Case  I. 
Take  ML  =f,  and  the  trapezoid  MLSN  =  R  (Fig.  227). 

The  pressure  on  the  bed  is  now  represented  by  the  trape- 
zoid MLSN. 


G 


.'.R  =  ^{ML-\-NS)ALY 
=  l{f-\-NS)t. 


1 ^i-^ — n=---, 


and 


M 


o 

Fig.  347. 


a 


2R 

ML-\-NS=-~ 


^s=lf-. 


The  ordinate  through  the  centre  of  gravity  of  the  trapezoid 
parallel  to  Zy>/cuts  MN  in  the  centre  of  pressure  F. 

Draw  ST  parallel  to  NM. 

The  moment  of  MLSN  with  respect  to  0 

=  moment  of  MTSN  with  respect  to  O 
-j-  moment  of  ZST^with  respect  to  O, 


or 


TS' 


^ML  +  NS)MN  .OF  =  o^^LT-^ 

:.--jtqt  =  l{ML-NS)-^^  =-[2f.-^)-^, 

Hence,         q  —  -A-^  —  i),  and  is  evidently  <  ^.   .     .     . 


(54) 


RE  TAINING-  WALLS. 


263 


Now  W  must  be  ;i  function  of  x,  the  vertical  depth  of  N 
below  B ;  P  also  may  be  a  function  of  x. 

Hence  if /is  given,  and  the  corresponding  value  of  q  from 
(53)  o""  (54)  substituted  in  (52),  x  may  be  found. 

When  (53)  is  employed,  the  value  of  x  found  must  make 

q>\' 

When  (54)  is  employed,  the  value  of  x  found  must  make 

q<\' 

Example.  The  rectangular  wall  in  [a),  the  safe  crushing 

strength  of  the  material  being  10,000  lbs.  per  square  foot(=/). 


By  (53), 


Substituting  in  (52), 

Hence, 

X  ^  9.03  ft. 


Again,  ^  > '■ —  >  .4248,  and  is  h  fortiori  >  ^. 


R 

=  W= 

:  500;jr 

Q 

I 
~  2 

X 

120* 

If  (54)  is  employed, 


Hence,  by  (52), 


1/80         \ 
^  =321— -ij. 


<  i 


^^ 


^ 


I  ,     » 


By  trial  x  is  found  to  lie  between  12  and  13 ;  each  of  these 
values  makes  ^  >  ^,  which  is  contrary  to  (54). 

The  first  is  therefore  the  correct  substitution. 

(c)  The  angle  between  the  directions  of  the  resultant  pres- 
sure and  a  normal  to  the  bed  must  be  less  than  the  angle  of 
friction. 


■.:--ll-:r. 


.^% 


t 


-<» 


.o. 


IMAGE  EVALUATION 
TEST  TARGET  (MT-3) 


1.0 


I.I 


1.25 


M 

2.0 


1.8 


1-4    11.6 


'm 


m 


/A 


<^^ 


(? 


/ 


^« 


-^ 


Photographic 

Sciences 
Corporation 


33  WEST  MAIN  STREET 

WEBSTER,  NY    14580 

(716)  872-<503 


i 


264 


THEORY  OF  STRUCTURES. 


Let  0  be  the  angle  of  friction,  R  the  mutual  normal  pressure. 
Resolving  along  the  bed  and  perpendicular  to  it, 


Pcos  a-\-  fi  —  W^sin  a  <iR  tan  0 


and 


P sin  «  4-  /?  +  Wcos a  =  R; 


P  cos  a-\-  ti  ~  Wsm  a 
"  /» sin  a~+"^+  H^cos  a 

which  reduces  to 


<  tan  0, 


/'(cos  /?+«  cos  0— sin  fi-\-a  sin  0)>  W{s\n  0  cos  a+cos  0sin  a), 
or 


/*  cos  /?  +  a  4-  <?^  <  W^sin  «  +  0, 


or 


or 


P{cos  fi  cos  a  +  0  —  sin  /?  sin  a  -|-  0)  <  Wsxn  a  +  0» 
Pcos/3 


.'.  tan 


-1 


Psin  /S  +  W^ 

P  cos  ft 
Psin  yS  +  fT 


<  tan  flf  +  0. 

-«<0 (55) 


17.  Rankine's  Theory  of  Earthwork  applied  to  Retain- 
C     ing-walls. — Fig.   228    represents  a 
vertical  section  of  a  wall  retaining 
earthwork.    AB  \s  a.  vertical  plane 
cutting  the  ground-surface  AC  in  the 
point  A. 
"^        Consider  the  equilibrium  of  the 
whole  mass  of  masonry  and  earth- 
work in  front  oi  AB.  ,        v 
Let  the  depth  ^^=  jr. 
The  total  prebsure  on  AB  is,  by 
(43). 


Pio.  aaS. 


-      1VX*        „  cos  6  —  Vcos'  6  —  cos'  0 

P  =  cos  a  7 . 

2  cos  <^  -|-  Vcos"  H  —  cos'  0 


LINE   OF  RUPTURE. 


265 


^      Its  point  of  application  is  D,  and  BD  =  — . 

Let  W  be  the  weight  of  the  whole  mass  under  considera- 
tion, and  let  its  direction  cut  the  base  of  the  wall  in  the  point  G. 
Let  F  be  the  centre  of  pressure  in  the  wall-base. 
Taking  moments  of  P  and  PF  about  F, 

P I  ^cos  ^  -  (^^+  j)  sin(6>+a)  |  t  W{qt  T  rt)  cos  a.  (56) 


The  other  conditions  of  equilibrium  may  be  discussed  as  in 
Art.  16. 

i8.  Line  of  Rupture. — Another  expression  for  the  press- 
ure on  AB  may  be  obtained  as  follows : 

If  the  whole  mass  in  front  of  AB  (Fig.  229)  were  suddenly 
removed,  some  of  the  earthwork  behind  AB  would  fall  away. 

Suppose  that  the  volume  ABC  would  slip 
along  the  plane  CB. 

The  stability  of  ABC  is  maintained  by  the 
reaction  P  ow  AB,  the  weight  W  oi  ABC,  and 
the  frictional  resistance  along  BC. 

Let  the  direction  of  P  make  an  angle  /3 
with  the  horizon. 

Let  the  angle  CBA  =  i. 

Let  R  be  the  mutual  pressure  on  the  plane  BC. 

Resolving  along  and  perpendicular  to  BC, 


Fig.  339. 


V:m.- 


and 


-/'cos (90° -»■  —  /?)+  H^cos«  =  ^tan0; 


P  sin  (90°  -/-/?)-(-  W^sin  i  -  R. 


im 


t 


.-.  -  P sin  {/3  +  t)  -f-  Wcos i  =  tan  0  | Pcos  {/3  +  t) -f  fTsin  t\ , 
and 


P=  W 


cos  i  —  sin  i  tan  0 


=  W-r 


cos  {i  +  0) 


sin  {fi  +  0  +  cos  (/«  +  0  tan  0  sin  {ti-\-i-\-  0)' 


iv 


t 


Ml' 


»i'l 


tr 


266 

Also 


THEORY  OF  STRUCTURES. 


BA.BC  .    . 
W=  w sin  { 


7vx^  COS  d  sin  t 
~2~  cos  {e -\- iy 


\P- 


zvx'  COS  ^  sin  t     cos  {t  -\-  0) 


2    cos {tf-{-t) sin {p-\-i-{-<p)' 


is?) 


!v;i 


The  only  variable  upon  which  P  depends  is  the  angle  i. 

Differentiating  the  right-hand  side  of  eq.  57  with  respect  to 
t  and  putting  the  result  equal  to  zero,  a  value  of  i  is  found  in 
terms  of  /?,  0  and  <p,  which  will  make  Pa.  maximum. 

The  line  inclined  at  this  angle  to  the  vertical  is  called  the 
line  of  rupture. 

If  the  ground-surface  is  horizontal,  ^  =  o. 

If  the  face  retaining  the  earth  is  vertical,  and  if  it  is  also 
assumed  that  the  friction  between  the  face  and  the  earthwork  is 
nil,  Pis  horizontal  and  fi  =  0.     Hence  (57)  becomes 


P- 


wx 


tan  i  cot  {i  +  0). 


(58) 


HI 


n 


' 


This  is  a  maximum  when  2i  =  90°  —  0,  and  then 


P- 


or 


7i'X 


tan(45-j)cot(45  +  2-j=  — 


,/  I  —  tan 


I  +  tan 


P- 


wx*  I  —  sin  0 
2    I  -[-  sin  0  ' 


.     .     .     .     (59) 


the  same  result  as  that  obtained  by  Rankine's  theory. 


PRACTICAL   RULES. 

The  following  is  an  easy  geometrical  proof  of  eq.  (59); 

On  any  line  KL  (Fig.  230)  de- 
scribe a  semicircle.  ^, . 

Draw  ATJ/ inclined  at  the  angle       //      v5< 
0  to  KL,  and  KN  inclined  at  the 
angle  i  to  KM. 

Join  NL,  cutting  KM  in  T. 

Let  0  be  the  middle  point  of  ^ 
the  arc  KM. 

Join  OL,  cutting  A'J/in  Y. 

Draw  NV  parallel  to  /i:.'!/.    Then 


267 


i\ 


Fig.  J30. 


,.  ,    ^,       NT     KN     NT      VY 
tan.cot(.  +  <?^)  =  ;^-^X-vZ-  =  ^^-=P^. 


VY 
The  ratio  7^   is  evidently  a  maximum  when  N  coincides 

with  O,  and  hence  tan  i  cot  {i  -\-  0)  is  a  maximum  when  A'.V 
coincides  with  KO. 

Now  the  arc  OK  =  the  arc  OM,  and  hence  the  angle  OKM 
—  the  angle  C'ZA'. 

Hence,  if  OKM  —  i,  OLK  must  also  =  i. 

But  OKL  +  6>ZAr  =  90°  =  /  4-  0  +  /  =  2?  4-  0. 


^  =  45    - 


0 


19.  Practical  Rules. — When  the  surface  of  the  earthwork 

is  horizontal  and  the  face  of  the  wall  against  which  it  abuts  ver- 
tical, the  pressure  on  the  wall  according  to  Rankine's  theory  is 


P=--^ 


sin  0 


2     I  +  sin  0  ' 

and  the  direction  of  P  is  horizontal. 

This  result  is  also  identical  with  that  obtained  in  Art.  18,  on 
the  assumption  of  Coulomb's  wedge  of  maximum  pressure 
(Poncelet's  Theory). 


Pit 


»4i::nl 


:.t,.  J 


268 


THEORY  OF  STRUCTURES. 


Experience  has  conclusively  proved  that  the  theoretical  value 
of  P  given  above  is  very  much  greater  than  its  real  value,  so 
that  the  thickness  of  a  wall  designed  in  accordance  with  theory 
would  be  in  excess  of  what  is  required  in  practice.  In  the 
deduction  of  the  formula,  indeed,  the  altogether  inadmissible 
assumption  is  made  that  there  is  no  friction  between  the  earth- 
work and  the  face  of  the  wall.  This  is  equivalent  to  the  sup- 
position that  the  face  is  perfectly  smooth  and  that  therefore 
the  pressure  acts  normally  to  it.  Boussinesque,  Levy,  and  St. 
Venant  have  demonstrated  that  the  hypothesis  of  a  normal 
pressure  only  holds  true, 

either,  first,  if  the  ground  surface  is  horizontal  and  the  wall- 

0 
face  inclined  at  an  angle  of  45° to  the  vertical, 

or,  second,  if  the  wall-face  is  vertical  and  the  ground-surface 
inclined  at  an  angle  0  to  the  horizon. 

When  the  surface  of  the  ground  is  horizontal  and  the  face 
of  the  wall  vertical,  and  when  0  =  45°,  the  above  formula  gives 
the  correct  magnitude  of  P.  Its  direction,  however,  is  not  hori- 
zontal, but  makes  an  angle  with  the  vertical  equal  to  the  r  ngle 
of  friction  between  the  earth  and  the  wall.  The  wall-face  is  gen- 
erally sufficiently  rough  to  hold  fast  a  layer  of  earth,  and  in  all 
probability  Boussinesque's  assumption  that  the  friction  between 
the  wall  and  the  earth  is  equal  to  that  inherent  in  the  earth  is 
a  near  approximation  to  the  truth.  The  direction  of  P  will 
thus  be  considerably  modified,  leading  to  a  smaller  moment  of 
stability  and  a  corresponding  diminution  in  the  necessary  thick- 
ness of  the  wall. 

In  practice  the  thrust  P  may  always  be  made  small  by 
carrying  up  the  backing  in  well-punned  horizontal  layers. 

In  order  to  neutralize  the  very  great  thrust  often  induced 
by  alternate  freezing  and  thawing  and  the  consequent  swelling, 
a  most  effective  expedient  is  to  give  a  batter  of  about  i  in  i  to 
the  rear  line  of  the  wall  extending  below  the  line  to  which  frost 
penetrates. 

The  greatest  difficulty  in  formulating  a  table  of  earth-thrusts 
arises  from  the  fact  that  there  is  an  infinite  variety  of  earth- 
work.    As  an  example  of  this,  Airy  states  that  he  has  found 


PRACTICAL  RULES. 


269 


the  cohesive  power  of  clay  to  vary  from  168  to  800  pounds  per 
square  foot,  the  corresponding  coefficients  of  friction  varying 
from  1. 1 5  to  .36,  and  that  even  this  wide  range  is  less  than 
might  be  found  in  practice. 

A  correct  theory  for  the  design  of  retaining-walls  is  as  yet 
wanting.  According  to  Baker,  experience  has  shown  that  with 
good  backing  and  a  good  foundation  the  stability  of  a  wall 
will  be  insured  by  making  its  thickness  one-fourth  the  height, 
and  giving  it  a  front  batter  of  i  or  2  in.  per  foot,  and  that  under 
no  conditions  of  ordinary  surcharge  or  heavy  backing  need  its 
thickness  exceed  one-half  the  height.  Baker's  usual  practice  in 
ground  of  average  character  is  to  make  the  thickness  one-third 
the  height  from  the  top  of  the  footings,  and  if  any  material  is 
taken  out  to  form  a  face  panel,  three-fourths  of  it  is  put  back 
in  the  form  of  a  pilaster. 

General  Fanshawe's  rule  for  brick  walls  of  rectangular 
section  retaining  ordinary  material  is  to  make  the  thickness 

24^  of  the  height  for  a  batter  of  i  in    5  ; 


25" 

•• 

(( 

"   I  in    6 ; 

26" 

« 

"   I  in    8  ; 

27" 

<< 

"   I  in  10; 

28" 

(( 

"   I  in  12 ; 

30" 

l( 

"   I  in  24; 

32" 

li 

for  a  vertical  wall 

The  thickness  at  the  footing  adopted  by  Vauban  for  walls 
with  a  front  batter  of  i  in  5  or  i  in  6  and  plumb  at  the  rear,  is 
approximately  given  by  the  empirical  formula 

thickness  =  .19// +  4  f^., 

H  being  the  height  of  the  wall  above  the  footing.  Counter- 
forts were  introduced  at  intervals  of  15  feet  for  walls  above  35 
feet  in  height,  and  at  intervals  of  12  feet  for  walls  of  less  height. 

The  counterfort  projects  from  the  wall  a  distance  of  -r  "i~  3  ^'• 

approximately,  and  the  approximate  width  of  the  counterfort  is 

-   +  3  ft.,  diminishing  to  --  +  2  ft. 


If,!;' 


f1    i.*<'l 


'  !  : 


m 


I  I 


ti  I 


)«'( 


270 


THEORY  OF  STRUCTURES. 


'  ■    II 


a?  J 


i>- 


*  i'  ^ 


Brunei  curved  the  face  of  the  wall  and  made  its  thickness 
one-fifth  or  one-sixth  the  height.  Counterforts  2  ft.  6  in.  in 
thickness  were  introduced  at  intervals  of  10  ft. 

The  vast  importance  of  the  foundation  will  be  better  appre- 
ciated by  bearing  in  mind  that  the  great  majority  of  failures 
have  been  due  to  defective  foundations.  If  water  can  percolate 
to  the  foundation,  a  softening  action  begins  and  a  consequent 
settlement  takes  place,  which  is  most  rapid  in  the  region  sub- 
jected to  the  greatest  pressure,  viz.,  the  toe.  In  order  to  coun- 
teract this  tendency  to  settle,  the  toe  may  be  supported  by  rak- 
ing piles,  the  rake  being  given  to  diminish  the  bending  action  of 
the  thrust  on  the  piles.  It  is  also  advisable  to  distribute  the 
weight  as  uniformly  as  possible  over  the  base,  a  condition  which 
is  not  compatible  with  large  front  batters  and  deep  offsets,  as 
they  tend  to  concentrate  weight  on  isolated  points.  In  the 
case  of  dock-walls,  too,  a  large  front  batter  will  keep  a  ship 
farther  away  from  the  coping  and  will  necessitate  thicker 
fenders,  as  well  as  cranes  with  wider  throws.  As  an  objection 
to  offsets  Bernays  urges  that,  in  settling,  the  backing  is  liable 
to  hang  upon  them,  forming  large  holes  underneath.  He  there- 
fore favors  the  substitution  of  'a  batter  for  the  offsets.  On  the 
other  hand,  if  water  stands  on  both  sides  of  the  walls,  the 
hydrostatic  pressure  on  the  offsets  will  greatly  increase  its 
stability. 

Dock-walls  are  liable  to  far  greater  variations  of  thrust  than 
ordinary  retaining-walls.  The  water  in  a  dock  with  an  im- 
permeable bottom  may  stand  at  a  much  higher  level  than  the 
water  at  the  back  of  the  wall,  and  its  pressure  may  thus  even 
more  than  neutralize  the  thrust  due  to  the  backing.  With  a 
porous  bottom  the  stability  of  a  wall  may  be  greatly  dimin- 
ished by  an  upward  pressure  on  the  base.  The  experience  of 
dock-wall  failures  has  led  to  the  conclusion  that  a  large  moment 
of  stability  is  not  of  so  much  importance  as  "  weight  with  a 
good  grip  on  the  ground."  Many  authorities,  both  practical 
and  theoretical,  have  urged  the  great  advantages  in  economy 
and  strength  attending  the  employment  of  counterforts.  The 
use  of  Portland  cement,  or  cement  concrete,  will  guard  against 
the  breaking  away  of  the  counterforts  from  the  main  body  of 


RESERVOIR    WALLS. 


271 


B  0 


the  wall,  as  has  often  happened  in  the  case  of  the  older  walls. 
But  a  uniform  distribution  of  pressure  as  well  as  of  weight  is 
important,  .ind  it  therefore  seems  more  desirable  to  introduce 
the  extra  weight  of  the  counterforts  into  the  main  wall.  Be- 
sides, the  building  of  the  counterforts  entails  of  itself  an  in- 
creased expense. 

20.  Reservoir  Walls. — Let /be  the  maximum  safe  press- 
ure per  square  foot  of  horizontal  base,  at  inner  face  of  a  full 
reservoir,  at  outer  face  when  empty. 

Let  w  be  the  wciglit  of  a  cubic  foot  of  the  masonry. 

Assume  that  the  wall  is  to  be  of  uniform  strength,  i.e.,  that 
the  section  of  the  wall  is  of  such  form  that 
in  passing  from  any  horizontal  section  to  the 
consecutive  one  below,  the  ratio  of  the  in- 
crement of  the  weight  to  the  increment  of 
the  surface  is  constant  and  equal  to/. 

Let  AB,  Fig.  231,  be  the  top  of  the  wall. 
Take  any  point  0  as  origin,  and  the  vertical 
through  O  as  the  axis  of  x. 

Let    OA  =  t„    0B  =  t^,  and  let 

T  =  t,  +  t,  =  AB. 

For  the  profile  AP  consider  a  layer  of  thickness  dx  at  a 
depth  X.    Then 

tvydx 


C/itN    G   (/j 
Fig.  231. 


dy 


=  /. 


(I) 


or 


w  y 


c  being  a  constant  of  integration. 
When  .a;  =  o,  /  =  /, ;      - 

/ 


o  =  —  log*  'i  +  <^» 


w 


X- '  *• 


% 


t   I 


.r^-J    ■>: 


I 


y 


is' 
11; 

I 


If 


II 


272 

and  hence 


THEORY  OF  STRUCTURES. 


f       y 


■      (2> 


which  is  the  equation  to  ^/*and  is  the  logarithmic  curve. 
It  may  be  similarly  shown  that  the  equation  to  BQ  is 


-  =  ^-log.f 


w 


•         •  • 


.     .    (3) 
Equations  (2)  and  (3)  may  also  be  written  in  the  forms 


y  -  t,ef 


and 


TO 
—-X 


y  -  V^ 


(4) 


(5) 


Corresponding  points  on  the  profiles,  e.g.,  /*and  Q,  have  a 
common  subtan^ent  of  the  constant  value  — ,  for 


NT  =  PNt.nNPT[=y^)={^ (6) 

Area  PNOA  =  />.  =  /,(/-> -  ^)  =  £(V^  -  /.),  (7) 
where  PN-  F,. 

Area  QNOB  =  J^'^ydx 

where  QN  =  V,. 

.-.  Area  QPAB  =  £-(F.+  K,  -7^^)  =  ^(7"  -  T),    .    (9) 
where  PQ  =  F,  +  F,  =  r. 


=  ~(I^,-a   (8^ 


RESERVOIR    WALLS. 


273 


Thus  the  area  of  the  portion  under  consideration  is  equal 
to  the  product  of  the  subtangent  and  the  diflference  of  thick- 
ness at  top  and  bottom. 

Lines  of  resistance  with  reservoir  empty.  Let  g^  be  the 
point  in  which  the  vertical  through  the  C.  of  G.  of  the  portion 
UAPN  intersects  PN.    Then 


ii 


Mm 


,rH'l  li'if' 


Ng,  X  area  OAPN  —  f  ydx^-  ; 
.*.  Ngl^  y^  -  A)-  =  r  -I     ydy  =  7  -  ( Y,'  -  /,») ; 


.•.^^,  =  -^4^' 


So  if  ^,  be  the  point  in  which  the  vertical  through  the 
C.  of  G.  of  the  portion  C^^QA'' intersects  QN, 


Ng,=. 


v^  +  f. 


Let  G  be  the  point   in  which  the  vertical    through   the 
C.  of  G.  of  the  whole  uiass  ABQP  intersects  PQ.     Then 

NG  X  area  ABQP  =  Ng,  X  area  A  ONP-  Ng^  X  area  BONQ, 


or 


NG^-f^  f;  -  /. + y.  -  0  =  7  -( y"  -  A')  -  -  --( y:  -  /,'). 


w 


i 


t.SfJR' 


56S: 


274 


THEORY  OF  STRUCTURES. 


The  horizontal  distance  between  G  and  a  vertical  t'lrough 
the  middle  point  of  AB 


=NG-\{t-Q= 


(K.-/.y-(K-0'_(K.~0-(K.-0 
4(K.-/.+  K-0-  4 


=  one  half  of  the  horizontal  distance  between  the  verticals 
through  the  middle  points  of  AB  and  CD. 

The  locus  of  G  can  therefore  be  easily  plotted. 

Lines  of  Resistance  zvith  Reservoir  Full. — Let  R  be  the  centre 

of  resistance  in  PQ  (Fig.  232). 

Draw  the  vertical  QS,  and  consider 
the  equilibrium  of  the  mass  QSAPQ. 

Let  w'  =  weight  of  a  cubic  foot  of 
water. 


Si       BOA 


w'x'  X 


Fig.  333. 


or 


w'x' 


-  =  moment  of  water-pressure 

against  QS  about  R 
=  moment  of  weight  of  QBS 
about   R  -f-  moment    of 
weight  of  QPAB  about  R, 


f. 


,     —  moment  of  QBS  about  R-\--\T'  ~  T)w.GR. 


The  first  term  on  the  right-hand  side  of  this  equation  is 
generally  very  small  and  may  be  disregarded,  the  error  being 
on  the  safe  side. 

In  such  case 

^„       I  vj'      x'' 
GR  =  ^ 


6  /  T'-  T' 
Also  the  mean  intensity  of  the  vertical  pressure 
_      _  TV  X  area  APQB  _  J    _T\ 

—A—  PQ  —jy      T'J' 


Is 
m 


RESERVOIR    WALLS. 

and  the  maximum  intensity  of  the  vertical  pressure 


275 


=  A  = 


2R 


{i-Z9)T 


^  =  1/ 


I  —  2q 


or 


=  y7(l-f-6y)=/(l+6^)[l   -y,j. 

General  Case. — Let  the  profile  be  of  any  form,  and  consi'  er 
.-iny  portion  ABQP,  Fig.  233. 

Take  the  vertical  through  Q  as  -"i 
the  axis  of  x,  and  the  horizontal 
line  coincident  .\/l.  top  of  wallas 
the  axis  of  J/. 

The  horizontal  distance  {y)  be- 
tween the  axis  of  x  and  the  vertical 
through  the  C.  of  G.  of  the  portion 
under  consideration  is  given  by  the 
equation 


■^'/'^^-^  ^  / V-^' 


QL-i-_-_- 


/  being  the  width,  dx  the  thickness,  ^^'^^  '33- 


and  y  the  horizontal  distance  from  OQ  of  the  C.  of  G.  of  any 
layer  AIN  at  a  depth  x  from  the  top. 

When  the  reservoir  is  empty,  the  deviation  of  the  centre  of 
resistance  from  the  centre  of  base 


=  gT=Y--y<-^ 


When  the  reservoir  is  full,  let  q'T  be  the  deviation  of  the 
centre  of  resistance  from  the  centre  of  the  base,  and  disregard 
the  moment  of  the  weight  of  the  water  between  OQ  and  the 
profile  BQ.     Then 


- 1 

,11 


276 


THE'.RY  OF  STRUCTURES. 


,^      moment  of  water  pr.  ±  moment  of  wt.  of  ABQP 

g  T=  rflZTf.-T-TTjTTT^ T   y 


=  ±6 


w'x' 


'i"^^-^ 


weight  of  ABQP 
±y^  Y. 


w 


Hence 


{g±<l')T  =  :^ 


tv'x* 


rv  I    tdx 

t/o 


> 


Fig.  234. 


21.  General  Equations  of  Stress. — Let  x,  y,  s  be  the 
co-ordinates  with  respect  to  three  rectanguhir  axes  of  any  point 

O  in  a  strained  body. 

Consider  the  equilibrium  of  an 
element  of  the  body  in  the  form 
of  an  indefinitely  small  parallelo- 
piped  with  its  edges  0A{=  dx), 
OB{--=.dy),  OC{=dz)  parallel  to 
the  axes  of  x,  y,  z.     It  is  assumed 

X   that  the  faces  of  the  element  arc 

sufficiently  small  to  allow  of  tlio 

distribution  of   stress  over  them 

being  regarded  as  uniform.     TIk" 

resultant  force  on  each  face  will 

therefore  be  a  single  force  acting  at  its  middle  point. 

Let  X^,  F, ,  Z^  be  the  components  parallel  to  the  axes  x,  y,  z 

of  the  resultant   force  per  unit  area,  on  the 

face  EC. 

"    X,,  Y^,  Z^  be    the    corresponding    components    for   the 

face^C. 
"    X^ ,  y, ,  Z,  be    the    corresponding    components    for   the 
face  ^5. 
These  components  are  functions' of  x,  y,  z,  and  therefore 
become 

-(^.+f4-(''.+'M-(^.+f4 

for  the  adjacent  face  AIJ\ 


ui^    mm 


GENERAL  EQUATIONS  OF  STRESS. 
dy"''r        V   '    Ty 


277 


-(-.+f4-(n+f4-(^.+f4 


for  the  adjacent  face  BD\ 

for  the  adjacent  face  DC. 
Hence,  the  total  stress  parallel  to  the  axis  of  4r 

=  X.dyds  -  [x,  +  '^-^^d^dydz  +  X,dzdx  -  [x^  +  "^-^^-d^dzdx 

J^X,dxdy-  \X,^'-j^dz]dxdy 


IdX.   ,  dX.  ,  dX\  ,    ,    , 


dy 

Similarly,  the  total  stress  parallel  to  the  axis  of  j^ 

(dV,   ,dY,      dY\ 
^~\'dx'^-dy'^'dz¥''^y'^'^ 

and  the  total  stress  parallel  to  the  axis  of  z 

IdZ,    .  dZ,  ,  dZX  .    ,   , 

Let  p  be  the  density  of  the  mass  at  O,  and  let  P,,  Py,  P,  be 
the  components  parallel  to  the  axes  of  x,  y,  z  of  the  external 
force,  per  unit  mass,  at  O. 

pdxdydzP^  is  the  component  parallel  to  the  axis  of  x  of 

the  external  force  on  the  element ; 
pdxdydzPy'xs  the  component  parallel  to  the  axis  of  _;/ of 

the  external  force  on  the  element ; 
pdxdydzP,  is  the  component  parallel  to  the  axis  of  z  of 
the  external  force  on  the  element. 


» .  'I 


278  THEORY  OF  STRUCTURES. 

The  element  is  in  equilibrium. 


dx 


dV,   ,  dV,  ,  dV, 
dx    ^    dy    '    dz 


dZ^       dZ.^ 
dx        dy 


pPy\ 


dz       '^  '' 


(I) 


These  are  the  general  equations  of  stress. 

Again,  take  moments  about  axes  through  the  centre  of  the 
element  parallel  to  the  axes  of  co-ordinates,  and  neglect  terms 
involving  {dxydydz,  dx{dyYdz,  dxdy{dz)^. 


V,  =  Z,,    Z,  =  X,,    and    X,=  F,. 


(--) 


Adopting  Lamp's  notation,  i.e.,  taking 
iV, ,  iVj ,  iV,  as  the  normal  intensities  of  stress  at  0  on  planes 
perpendicular  to  the  axes  of  x,  y,  z  ; 
T^  as  the  tangential  intensity  of  stress  at  O  on  a  plane 
perpendicular  to  the  axis  of  x  if  due  to  a  stress 
parallel  to  the  axis  of  y,  or  on  a  plane  perpen- 
dicular to  the  axis  of  y  if  due  to  a  stress  parallel 
to  the  axis  of  x  ;  and  T^ ,  7",  similarly, — equa- 
tions (i)  become 


dx"^  dy"^  dz  -^P^-^' 


y  ' 


dx  ^  dy  ^  dz  ~P  ' 


dx  "^  dy  ~^  dz  ~  f    ' 


.    i3) 


GENERAL  EQUATIONS  OF  STRESS. 


279 


Next   consider  the  equilibrium  of   a  tetrahedral  element 
having  three  of  its  laces  parallel  to 
the  co-ordinate  planes.     Let  /,  w,  n 
be  the  direction-cosines  of  the  normal 
to  the  fourth  face. 

Also,  let  X,  V,  Z  be  the  compo- 
nents parallel  to  the  axes  of  x,  y,  z 
of  the  intensity  of  stress  R  on  the 
fourth  face. 

X=lN,-\-  w.  r,  +  «  r,  +  \pPJdx. 

But  the  last  term  disappears   in  i^'g-  »3s. 

the  limit  when  the  tetrahedron  is  indefinitely  small,  and  hence 


Zz=lT,-\-mT,  -\-nN,. 


(4) 


These  three  equations  define  R  in  direction  and  magnitude 
when  the  stresses  on  the  three  rectangular  planes  are  known. 

Let  it  be  required  to  determine  the  planes  upon  which  the 
stress  is  wholly  normal.     We  have 


X  =  IR,     Y=  inR,    Z  =  nR.  . 


(5) 


Substituting  these  values  of  X,  Y,  Z  in  eqs.  (4)  and  eliminat- 
ing /,  m,  n,  we  obtain 

-{N,N,N,  -  N,T:  -  N,T:-N,T:-\-2T,T,T,)=0',   (6) 

a  cubic  equation  giving  three  real  values  for  R,  and  therefore 
three  sets  of  values  for  /,  in,  and  «,  showing  that  there  are  three 
planes  at  O  on  each  of  which  the  intensity  of  stress  is  wholly 
normal.  These  planes  are  at  right  angles  to  each  other  and 
are  c^WqA  principal  planes,  the  corresponding  stresses  hemg  prin- 
cipal stresses.     They  are  the  principal  planes  of  the  quadric, 

N,x'  +  N,f  +  A^,j'  +  2  T,yz  +  2  T,2x  +  2  T,xy  =  c.      (7) 


I  \  \ 


i.       ! 


"  ^iiii 


lilST 


H 


im: 


;l^- 


280 


THEORY  OF  STRUCTURES. 


For,  the  equation  to  the  tangent  plane  at  the  extremity  of  a 
radius  r  whose  direction-cosines  are  /, ;«,  n  is 


Xrx  ■\-  Yry  -\-  Zrs  =  c, 


(8) 


and  the  equation  of  the  parallel  diametral  plane  is 


Xx -\-  Yy  ■\- Zz  —  o (9) 

The  direction-cosines  of  the  perpendicular  to  this  plane  are 


X__        K_         f_ 


so  that  the  resultant  stress  R  must  act  in  the  direction  of  this 
perpendicular. 

Hence  the  intensities  of  stress  on  the  planes  perpendicular 
to  the  axes  of  the  quadric  (7)  are  wholly  normal. 

Refer  the  quadric  to  its  principal  planes  as  planes  of  refer- 
ence.   All  the  7"s  vanish  and  its  equation  becomes 


Also,  the  general  equations  (3)  become 


(10) 


dy        *^  '' 


dN, 
de 


=  pP,- 


•         •        • 


.    .    .    (II) 


Again, 


(ir+(0+©'='-+"''+''-= 


I.  .  .  (12) 


M^ 


IV 


iittx 


RELATION  BETWEEN  STRESS  AND  STRAIN. 


281 


Consider  X,  Y,  Z  as  the  co-ordinates  of  the  extremity  of 
the  straight  line  representing  R  in  direction  and  magnitude. 
Equation  (12)  is  then  the  equation  to  an  ellipsoid  whose  semi- 
axes  are  iV, ,  iV, ,  TV, .  As  a  plane  at  O  turns  round  (?  as  a  fixed 
centre,  the  extremity  of  a  line  representing  the  intensity  of 
stress/?  on  the  plane  will  trace  out  an  ellipsoid.  This  ellipsoid 
is  called  the  ellipsoid  of  stress. 

Note  I.  The  coefficients  in  the  cubic  equation  (6)  are  in- 
variants. Thus,  N^-\-  N^-\-  N,  is  constant,  or  the  sum  of  three 
normal  intensities  of  stress  on  three  planes  placed  at  right 
angles  at  any  point  of  a  strained  body  is  the  same  for  all 
positions  of  the  three  planes. 

Note  2.  The  perpendicular/  from  O  on  the  tangent  plane, 
equation  (8), 


=  Wr=^- 


pr  ^  ^' 


Note  3.  Let  the  stress  be  the  same  for  all  positions  of  the 
plane  at  0.  Then  N,—N^  =  N,,  and  the  ellipsoid  (12)  be- 
comes a  sphere.  The  stress  is  therefore  everywhere  normal, 
and  the  body  must  be  a  perfect  fluid.  Conversely,  if  the 
stress  is  everywhere  normal,  the  body  must  be  a  perfect  fluid, 
the  ellipsoid  becomes  a  sphere,  and  therefore  iV,  =  iV,  =  TV, . 

22.  Relation  between  Stress  and  Strain.— In  Art.  13  it 
was  shown  that  when  the  size  and  figure  of  a  body  are  altered 
in  two  dimensions,  there  is  an  ellipse  of  strain  analogous  to  the 
ellipse  of  stress.  If  the  alteration  takes  place  in  three  dimen- 
sions, it  may  be  similarly  shown  that  every  state  of  strain  may 
be  represented  by  an  ellipsoid  of  strain  analogous  to  the  ellip- 
sold  of  stress.  The  axes  of  the  ellipsoid  are  the  principal  axes 
of  strain,  and  every  strain  may  be  resolved  into  three  simple 
strains  parallel  to  these  axes. 


i 

■n 

I 

if 

11' 

W 

ml 

R 

il 

W 

11 

:-r;  \m 


I;-!,";'! 


■If 


!•  •;• 


282         .  THEORY  OF  STRUCTURES. 

It  is  assumed  that  the  strains  remain  very  small,  that  the 
jj  stresses  developed  are  proportional 

S' 


to   the    corresponding    strains,   and 

that  their  effects  may  be  superposed. 

Consider  an   element  of  the  uii- 

'_ [_.Q'      strained  body  in  the  form  of  a  rcct- 

"*  angular    parallelopiped,    having    its 
edges  PQ  (=  h),  PR  (=  k\  PS  {=  I) 
parallel  to  the  axes  of  co-ordinates. 
When  the  body  is  strained,  the 
^'"'  '^*'  element  becomes  distorted,  the  new 

edges  being  P'Q',  P'R',  P'S'. 

Let  X,  y,  z  be  the  co-ordinates  of  P. 

Let  X  -\-  u,  y  -\-  V,  z  -\-  w  ho.  the  co-ordinates  of  P'. 

By  Taylor's  Theorem  the  co-ordinates  with  respect  to  /*'  of 

„,         ,/     ,   dti\      ,dv       .dtv 
ay         \        ay  I        ay 


t  >  • 


(14) 


Hence,  strain  parallel  to  axis  of  j»r  = 


y  = 


z  = 


p'Q'-PQ  du:\ 

PQ  ~  ~dx' 

P'R' -PR    dv  ^ 

PR  '   ~dy' 

P'S' -PS    dw 
PS       ~dz' 


(•5) 


ISOTROPIC  BODIES. 


285 


Again,  cos  QP'R' 

du 
dxidy 


(■+S)^+('+-)*+ 


dw  dw 
dy  I  dx   '    dy  dx 


[l-+(Sv©v©'H(^p"+(.-4;)vgf)'l]* 

In  the  limit,  this  reduces  to 


•     I     •     • 


(16) 


cos  G'/"/e' =  ^  +  $^^ 

dy       dx 
Similarly,  cos  QP'S'  =  ~  ^~\V 

C0SR'P'S'=:^  +  '^. 

ay       dz  ^ 
Volume  of  unstrained  element  =  hkl\ 

Volume  of  distorted  element  =  ^^'^A  ^+77")\^+3~)(  ^+7") 

multiplied  by  the  cosines 
of  small  angles 

in  the  limit. 
Difference  of  volume  dii    ,  dv    .  dtv  ,    , 

^•^ . . .  3:::^ L  ^       ^       ^      (17) 

"  Vol.  of  unstrained  element       dx       dy       dz^   '     '     ' 

=  the  volume  or  cubic  strain. 

23.  Icc-*Topic  Bodies,  i.e.,  bodies  possessing  the  same  elas- 
tic properties  in  all  directions. 

A  nornja)  stress  of  intensity  TV,  parallel  to  the  axis  of  x 

N 
produces   a  simple  longitudinal   strain  — i,  and   two   simple 

.V     ^ 

lateral  strains,  each  = ^,  parallel  to  the  axes  of  y  and  s, 

nth 


'*•, 


•i.!| 


'    t 


it  Lis 


>  I  .  - 


x.miL'iii&i^ 


iiH^lli 


384 


THEORY  OF  STRUCTURES. 


E  being  the  ordinary  modulus  of  elasticity  and  — ,  Poisson's 


/« 


ratio  (Art.  3,  Chap.  III). 

Normal  stresses  N^ ,  N^  parallel  to  the  axes  o{  y  and  2  may 
be  similarly  treated. 

Let  the  three  normal  stresses  act  simultaneously  and  super- 
pose the  results.     Then 


total  strain  parallel  to  axis  of  jr  =  -' 


E 


II.       II 


II        fi 


II 


II 


y  = 


inE        dx ' 

TV,    N,-\-N,    dv 


z  =  -=^ 


E         inE       dy 

N,    N,-\-N,_div 
mE    ~  dz 


:  )-    (18) 


lIP' 


V 


ill 


i 


The  form  in  which  these  equations  are  given  is  due  to 
Grashof. 

Solving  for  iV;,  TV,, -/V,, 


N. 
K 
N, 


m{m  —  \)E      du 


+  ; 


mE 


{m  -f-  i)(w—  2)  dx   "^  (;«  +  i)('«—  2)\dy 


Idv       dw\  ^ 


m{m  —  i)E     dv 


+  - 


mE  idw  ,  du 


(w-f-  i){m—  2)dy    '^(;«+  ^){^—  2)\d2 


^dxr 


Y  (19) 


_    m{m  —  i)E     dw 


vr  + 


mE 


Idu    .  dv 


{m  +  i)(w—  2)  dz    '  (w  4-  iXw—  2)\dx    '  dy 


^dvl' 


The  last  equations  may  be  written 


dx  \iy        dz  ]  ' 

.dv    ,    yidw   ,   du\ 


dy 
.dw 


=  <^  +  H.T  +  ^)^J 


.      .      (20) 


ISOTROPIC  BODIES. 


285 


where  \  = 


mE 


(;«  +  iX''^—  2) 


,  is  the  coefficient  of  dilatation,  and 


A  _       in{m—\)E 
~  {m-\-  \){tH—  2) 

Again,  the  straining  changes  the  angle  RPS  by  an  amount 

'-1^  4-  —  .   producing  two  tangential  stresses,  each   equal   to 
dy        as 

Gl  --  +  "T  )>  parallel  to  the  axes  of  ^  and  s. 

Similarly,  ^' =  ^('1  +  ll"')  '  r" i") 

^■  =  <f  +  '|)-. 

G  is  called  the  coefficient  of  rigidity  or  transverse  elasticity, 
and  is  designated  n  in  Thomson  and  Tait's  notation,  and  yw  in 
Lamp's  notation. 

Relation  betzveen  A,  A,  and  G. — Equations  (20)  and  (21)  pre- 
serve the  same  forms  whatever  rectangular  axes  may  be  chosen. 

Keep  the  axis  of  z  fixed  and  turn  the  axes  of  x  and  / 
through  an  angle  a. 

Let  N^  be  the  normal  stress  parallel  to  the  new  axis  of  x. 

:.  N^  —  TV,  cos'  a  -\-  N^  sin'  «  +  2  T'j  sin  or  cos  a.       (22) 

Let  x\y'  and  u',  v'  be  the  new  co-ordinates  and  displace- 
ments. 

„     dti   ,  dv  .  dw       ^       du'   .  dv'   ,  dw'  .         •        •     ^ 
For  ---{-  —  -}-  -~,  =  d  =  — -,  -f-  _-  -f  IS  an  mvanant. 

dx      dy      dz  dx       dy        dz 


Mill 


';  iff 


.AM  -J 


i 


ili 


ii 
i! 


ill    : 


286 


THEORY  OF  STRUCTURES. 


The  values  of  N^'  given  by  eqs.  (22)  and  (23)  must  be 
identical.     Now, 


X  =^  x'  cos  a  —  y  sin  a,    _;/  =  .ir'  sin  a  -|"  J''  cos  « ; 
ti'  =  «  cos  a  -\-  V  s\n  a,     v'  ■=.  —  u  sin  a  -f-  ^  cos  a , 

du'       du  .   dv     . 

.•.-  =  -cos«+-,s.n« 

du        ,       .  dv    .  ^       ,    /i/«    ,   rfz/\  . 
=  ~r-  cos  a  +  -7-  sin  «  +  I  -7-  +  -t-  jsin  a  cos  a 
</.r  dv  \dv       dxl 


I .  (24) 


du        .       .   dv    .  ^       I    ^3    • 
=  -T—  cos'  «  +  -7—  sin  «  +  — ?  sin  «  cos  a  ; 
f/.f  dy  G 

and  by  eq.  (23), 

N;  =  (^  -^)(^-  cos'  a+^  sin'  «+^  sin  a  cos  a)  +  ^6*.     (25) 

Also  by  eqs.  (20)  and  (22), 

N:  =  (A—X){^  cos'  a+^  sin  a  +4^  sin  a  cos  a]+A^.  (26) 
^dx  dy  A— A.  ,  / 


Eqs.  (25)  and  (26)  must  be  identical. 


.'.G  = 


A  -\ 


inE 


2  2{ni  -|-  i) 

Adding  together  equations  (20), 


-li^n.      .    .    (27) 


^.+^.+^.=(^  +  ^<+^j+f) 


dy 
inE   (du    ,  dv    ,  d'i 


Wi      -: 


Idu    \dv_.  dziA 
m  —  2^dx       dy       dzj 

It  may  be  easily  shown  that  the  normal  stresses  can  each 
be  separated  into  a  fluid  pressure/  and  a  distorting  stress. 


If! 


APPLICATIONS.  287 

Hence,  putting 

x{m  —  2)  \dx    '  dy       dzl 


the  cubic  elasticity  =  v j  , 

■'       du       dv       dw 

di'^d^^lh 


dy 
viE 


3(;«  -  2) 


=  K.  (28) 


24.  Applications.— I.  Traction. — One  end  of  a  cylindrical 
bar  of  isotropic  material  is  fixed  and  the  bar  is  stretched  in  the 
direction  of  its  length.  The  axis  of  the  bar  is  the  only  line 
not  moved  laterally  by  contraction. 

Take  this  line  as  the  axis  of  x. 

The  displacements  ti,  v,  w  of  any  point  x,y,  z  may  be  ex-- 
pressed  in  the  form 


u  ■=.  ax,     V  ■=  —  fiy,     w  =  —  /3s. 
By  eqs.  (20)  and  (29), 

N,=aA  —  2/?A 


.     .    (29) 


(30) 
(31) 


By  eqs.  (21)  and  (29),  all  the  tangential  stresses  vanish. 

Hence,  since  TV, ,  A^, ,  iV,  are  constant,  and  since  the  equa- 
tions of  internal  equilibrium  contain  only  differential  coeffi- 
cients of  the  stresses,  the  hypothesis,  eq.  (29),  satisfies  these 
equations. 

First.  Let  iV!,  =  o  =  iV, ;  i.e.,  let  no  external  force  act  upon 
the  curved  surface. 


or 


A 

+  A(/3  + 

ar)  =  0, 

a 

X 

A+Ji 

_  I 

m 

•         •         •         t 


(32) 


Thus,  the  coefficient  of  contraction  is  less  than  the  coeflficient 
of  expansion.    ' 


m 

1  lij; 
''    1 

1 

f 

■^    ■!11I 

-J 

Pi 

^\''t^'M 


■|'i*'l    ;! 


^^  i 


288  THEORY  OF  STRUCTURES. 

Again,  by  eqs.  (30)  and  (32), 


a 


a 


m 


(33) 


Second.  If  the  bar,  instead  of  being  free  to  move  laterally, 
has  its  surface  acted  upon  by  a  uniform  pressure  P,  then 


N,  =  iV,  =  P. 


By  eqs.  (31)  and  (32), 


AP—XN, 


a  -  \{N,  +  2P)  -  AN, 


(35) 


For  example,  let  Phe  sufficient  to  prevent  lateral  contrac- 
tion.    Then  fi  ■=  o  and,  by  eqs.  (31)  and  (35), 

AP 
aA  =  N,  =  -j-  =  (;«  -  i)P. 


( 


W 


■ 


i 
i- 


2.  Torsion. — {a)  Let  a  circular  cylinder  (hollow  or  solid)  of 
length  /  undergo  torsion  around  its  axis  (the  axis  of  x),  and  let 
t  be  the  angle  through  which  one  end  is  twisted  relatively  to 
the  other.     A  point  in  a  transverse  section  distant  x  from  the 

latter  will  be  twisted  through  an  angle  x-. 

The  displacements  11,  v,  w  of  the  point  x,y,  s  in  this  section 
may  be  expressed  in  the  form 


u  =  o,     V  =z  —  ZX-,     IV  =  -\-  }'X-  . 


By  eqs.  (20)  and  (21), 


and 


r,  =  o,    r,  =  +  Gyj ,  r,  =  -  Gz-  . 


APPLICATIONS. 


289 


The  algebraic  sum  of  the  moments  of  7!,,  7",  with  respect 
to  the  axis 

^ 


w 

IH 

n 

ffi 

w 

1 

w 

■iri 

nn 

\'h 

r  being  the  distance  of  the  point  {x,y,  z)  from  the  axis. 

Hence,    the  moment   J/,  —  Pp  (Chap.  IX),   of   the   couple 
producing  torsion 


=  G\f  r\iS  =  d-I  = 


cei, 


dS  being  an  element  of  the  area  at  {x,y,z),  I  the  polar  moment 
of  inertia,  and  ^  the  torsion  per  unit  of  length  cf  the  cylinder, 
or  the  rate  of  twist. 

The  torsional  rigidity  of  a  solid  cylinder 


R  being  the  radius  of  the  cylinder. 

{b)  Torsion  of  a  bar  of  elliptic  section. 

The  displacements  u,  v,  w  may  now  be  expressed  in  the 
form 

ti  =  F{y,  z),     V  =  —  0x2,     zv  =  6xy. 


dii 


—  o 


dv  _  dw 
dy       dz  ' 


dx 
Hence,  by  the  general  eqs.  (3), 


d''u    ,    d'li 

—  A =  o. 

df  ^  dz' 


(35) 


(36) 


■\\ii 


1  ?■ .  • 


t    :    ,,«>) 


<l         '91 


■      ■        U'V' 


1! 

HmHi  ' 

\m 


if 


AIbo,  the  BUifacc  Btiessen  f»ic  ?em; 


aiul  UctKC,  by  cqs.  ^^35), 


'.  ,/5  - ''f/j- =  f^^,/e  !-/,/»').   ....    ^38 1 


Tbis  o»j\iiUioti  \mi^t  hoUl  (riic  ;\t  Ihr  r^nilruo. 
Lot  Iho  cquiUion  to  llio  olliptiv  sc'cli»)n  l>o 


y     1   « 


»••*»•# 


1       •       >       •       •       •       • 


•  (40) 


aiul  by  cq.  (  uS\ 


(,|h 


n  =  f/i'ff  satisfies  this  hist  equation  ami  also  rq.  {^6),  if 


f^ 


(4.') 


Ajjaitu  the  algebraic  sum  of  the  moments  T, ,  7",  with  it 
spcct  to  the  axis  of  x. 


•    .    .    (43^ 


t^rrfrfrirfOf^x 


agt 


The  tulttl  iiiuinriil  {/if   «if  llir  cMUjilr  prudiicing  turilun 


f>'  4-  / 


/[,y  [-f>wus 


s=  r;/' 


fr/'' 


..I  < 


ami  flio  torsiiiiml  iljjiility 


/]/ 


nf> 


,,,■ 


J  =^  "  /,.-q:- 


!••  I  •  ■  •  «  •  ■  • 


(44) 


(t)  TnrNlmi  of  a  bur  nf  rrMniifiiiliU"  ^rrtlon. 
As  In  case  (A),  ff  nnist;  '•mtiflfy  the  n(|iiiitioti 


-^  o. 


•         •««••• 


(45) 


AUo,  the  iMUmtinii^  of  ('(»ii(li(i(»n  coiirspniKliii^,'  1(»  r«j,  (jH)  arr 
-4-.  «-  ^7  *-  0    wlipn    V  -*^  :h  (^ (4''') 


,iii<i 


'0* 


-\.(ij>-,o     when     /rrn  J:r;       ...     (47) 


.'/' iiiul  2r  ((^  <  f)  bcinf^  the  sidcR  of  thr  rcctanfrlc,     Thr  total 
Dininoiit  of  torsion,  v\/.,,^/'{7\j>  ■--  T^ayfS  is  then  found  lo  bo 


M 


r  *''"  1 

,     ,  Ian  //  (2n    I    l)    , 


3       rr*  r  '^" 


If  //  ex  If,  i.e.,  if  the  section  ib  a  square,  eq.  (48)  bi-comei 

il/a=  .843462 /67y (4(>) 

/(      lb*)  bcinpj  the  momrnt  of  inertia  with  respect  to  the  axes. 
lSf<-  Cha|).  IX). 


if 
■f 


;c| 


Li 


!i   I 


292  THEORY  OF  STRUCTURES. 

k'  •      •        .  ,  -  "        .....'. 

If  -  is  very  small,  eq.  (48)  becomes 

M=xU^cGe{^^-,2V>).,     ~   .    .     .    (50) 

The  torsional  rg'dity  of  a  rectangular  section  is  sometimes  ex- 
pressed by  the  formula  ....  . 

M  _  5      dY 
'    ;    -    .    ^  (^  -  isd'  +  c'        •    • ^50 

For  the  further  treatment  of  this  subject,  the  student  is  re- 
ferred to  St.  Venant's  edition  of  Clebsch,  and  to  Thomson  and 
Tait's  Natural  Philosophy. 

3.  Wor&  done  in  the  small  strain  of  a  body  (Clapeyron's 
Theorem). — M  ultiply  eqs.  (3 )  hyudxdy  dz,  v  dx  dy  dz,  ivdxdydz, 
and  find  the  triple  integral  of  theif  sum  throughout  the  whole 
of  the  solid. 

The  terms  involving  the  components  P^^P^^  P^  may  be  dis- 
regarded, as  the  deformations  due  to  their  action  are  generally 
inappreciable. 

Also,  ' 

'SSf-d^'''^^'^y^'  ■ 

=J'f^NJu,  -  NJ'uJ')dydz  -fffNf^dxdydz, 

NJ,  NJ'  being  the  values  of  N^  at  the  two  points  in  which  the 
line  parallel  to  the  axis  of  x  cuts  the  surface  of  the  body,  aiul 
uj,  uj'  the  corresponding  values  of  u. 

Let  dS,  dS'  be  the  elementary  areas  of  the  surface  at  these 
points,  and  /',  /"  the  cosines  of  the  angles  between  the  normals 
to  these  elements  and  the  axis  of  x. 

The  double  integral  on  the  right-hand  side  of  the  last  equa- 
tion then  becomes  _        .  .  ,       •    ■     ■       . 

//{NJl'uJdS  -  NJ'V'uJ'dS)  =  :S{NJudS). 


APPLICATIOITZ  293 

Treating  the  other  terms  similarly, 

0  =  2K^/+  T,in  +  T.,n)u  +  (7,/+  N,m  +  T,n)v 

+  ( r/  +  r.  w  +  -N,n)w  \  dS 


df^ 


.    ^  (dv       dzv\  Idw      du\  Idu    .  dv\] 

+  ^'  UJ  +^"J  +  ^'  i^  +  ^;  +  ^'  l^  +  ^J  f 

Hence,  the  ivork  done  =  ^2{Xu  -\-  Yv  -\-  Zw)dS 
=  \Sffdxdydz I  c~^^^{N.  +  ^,  +  ^,r        • 

=  ,fffd.dyd.{^^^^^     -. 


E  being  the  ordinary  modulus  of  elasticity. 


]• 


t» 


m 


tlie 
ami 

;hcsc 
mals 

:qua- 


MH: 


*v 


294 


THEORY  OF  STRUCTURES. 


EXAMPLES. 

1.  At  a  point  within  q  strained  solid  there  are  two  conjugate  stresses 
viz.,  a  tension  of  200  lbs.  and  a  thrust  of  150  lbs.  per  square  inch,  tlir 
common  obliquity  being  30".  Find  (a)  the  principal  stresses;  {b)  \.\\\i 
maximum  shear  and  the  direction  and  magnitude  of  the  correspond itij^ 
resultant  stress ;  (c)  the  resultant  stress  upon  a  plane  inclined  at  30'  to 
the  axis  of  greatest  principal  stress. 

Am. — {li)  A  tension  of  204.65  lbs.  and  a  thrust  of  146  95  lbs.  per  sq  in. 
(b)  175.8  lbs.  per  sq.  in.;    173.2  lbs.  in  a  direction   making  an 
angle  of  40°  13'  with  the  axis  of  greatest  principal  stress. 
(0  163.3  lbs.  per  sq.  in. 

2.  A  wall  with  a  plumb  rear  face  is  to  be  30  ft.  high  and  4  ft.  wide  at 
the  top  ;  the  earth  slopes  up  from  the  inner  edge  at  the  angle  of  20°, 
30°  being  the  angle  of  repose.  Assuming  Rankine's  theory,  determine 
the  proper  width  of  the  base,  the  masonry  weighing  144  lbs.  per  cubic 
foot,  and  tiie  eartli  no  lbs. 

3.  A  wall  6  ft.  wide  at  the  bottom,  plumb  at  the  rear,  and  with  a 
front  batter  of  i  in  12,  retains  water  level  with  the  top.  Find  {a)  the 
limiting  position  of  the  centre  of  pressure  at  the  base  so  that  the  stress 
may  be  nov/here  negative. 

How  {h)  high  may  the  wall  be  built  when  subjected  to  this  condition.' 
(a  cubic  foot  of  masonry  =  125  lbs.). 

Ans,  {a)  12  in.  from  middle  poMit  of  base;  (*)  height  =  8.9  ft. 

4.  A  wall  is  built  up  in  layers,  the  water  face  being  plumb  and  the 
rear  stepped.  If  t  be  the  thickness  of  the  «th  layer  and  y  the  depth  o( 
water  above  its  lower  face,  show  that  width  of  layer  x  thickness  of  layer 

=  V4'''''  +  (>Atz  4-  ;;//(''  —  2A  ;  .,'/  being  the  sectional  area  of  the  wall 
above  the  layer  in  question,  s  the  horizontal  distance  between  the  water 
face  and  the  Ime  of  action  of  the  resultant  weight  above  the  layer,  /  the 
layer's  thickness,  and  ;//  the  ratio  of  the  specific  weights  of  the  water  and 
masonry. 

5.  At  a  point  within  a  strained  solid,  the  stresses  on  two  planes  at 
right  angles  to  each  other  are  a  thrust  of  30 1^2  lbs.  and  a  tension  of  60 
lbs.  per  square  inch,  the  obliquities  being  45°  and  30°  respectively. 
Determine  (<»)  the  principal  stresses;  (/')  tiie  ellipse  of  stress;  (<)  ilie 
intensity  of  stress  upon  a  plane  inclined  at  60°  to  the  major  axis. 

Ans. — Or)  A  thrust  of  61.76  lbs.  and  a  tension  of  39.80  lbs. 
{c\  A  thrust  of  66.5  lbs. 


y  k 


EXAMPLES, 


295 


•I ' 


-\\- 


6.  If  the  principles  of  the  ellipse  of  stress  arc  applicable  within  a 
mass  of  earth,  and  if  at  any  point  of  the  mass  the  stress  upon  a  plane  is 
double  its  conjugate  stress,  the  angle  between  the  two  stresses  being 
20°  28',  show  that  the  angle  of  repose  of  the  earth  is  28'.  i. 

7.  The  total  stress  at  a  point  O  upon  a  plane  ///>  is  60  lbs  per  square 
in'''\  and  its  obliquity  is  30" ;  the  normal  component  upon  a  plane  CD 
at  the  point  O  is  4.0  lbs,  per  square  inch  ;  CD  is  perpendicular  to  AH. 
Find  ((«)  the  total  stress  upon  CD,  and  also  its  obliquity  ;  {h)  the  princi- 
pal stresses  at  O ;  {c)  the  equal  conjugate  stresses  at  O. 

Ans. — (rt)tan-'(});  50  lbs. 

{!))  76.57  lbs.  and  15.39  lbs. 

(r)  34.23  lbs, ;  obliquity  =  41°  42'. 

8.  Assuming  Rankine's  theory,  find  the  pressure  on  the  vertical  face 
of  a  retaining-wall,  30  ft.  high,  which  retains  earth  sloping  up  from  the 
top  at  the  angle  of  repose,  viz.,  30'. 

(Weight  of  masonry  =  128  lbs.  per  cubic  foot.;  weight  of  earth  =  120 
lbs.  per  cubic  foot.)  Ans.  46,764  lbs. 

9.  At  a  point  within  a  strained  solid  the  stress  on  one  plane  is  a  ten- 
sion of  50  lbs.  per  square  inch  with  an  obliquity  of  30",  and  upon  a 
second  plane  is  a  compression  of  150  !bs.  per  square  inch  with  an  ob- 
liciuity  of  45".  Find  (a)  the  principal  stresses ;  {b)  the  angle  between  the 
two  planes ;  (c)  the  plane  upon  which  the  resultant  stress  is  a  shear,  and 
the  amount  of  the  shear. 

Ans. — (a)  p\  r-  153.8  lbs.  (comp.) ;  /j  =  —  20  lbs.  (tens.) 
(d)  71"  55'. 
(<r)  86.88  ibs.;  y  =  19°  50'. 

10.  At  a  point  within  a  strained  solid  the  stress  on  one  plane  is  a  ten- 
sion of  100  lbs.  per  square  inch  with  an  obliquity  of  30',  and  on  a  second 
plane  a  compression  of  50  lbs.  with  an  obliquity  of  60°.  Find  (a)  the 
an^le  between  the  planes;  {/>)  the  plane  upon  which  the  stress  is  wholly 
a  shear;  (t)  the  planes  of  principal  stress. 

Ans. — (a)  1 1'  38'. 

(/>}  64.6  lbs  ;  y  =  3"  26'. 

(c)  />i  =  106  46  (tens  );/„  =  —  39.26  (compr.). 

1 1.  In  the  preceding  question  find  the  conjugate  stresses  at  the  given 
point  having  the  common  obliquity  45°.  Ans.  Impossible. 

12.  At  a  point  within  a  strained  mass  the  principal  stresses  at  a  given 
point  are  in  the  ratio  of  3  to  i.  Find  the  ratio  of  the  conjugate  stresses 
at  the  same  poit\t  having  the  common  obliquity  30".  Also  find  the  in- 
clination of  the  axis  of  greatest  principal  stress  to  the  horizontal. 

Ans.   Equal ;  60". 

13.  A  wall  3  feet  thick,  of  rectangular  section  and  weighing  125  lbs. 
\)v\-  cubic  foot,  is  subjected  to  a  horizontal  thrust  of  800  lbs.  per  foot  run 


296 


THEOKY  OF  STRUCTURES. 


il 


ill. 


f«i  ■ 


at  its  t(jp.  What  sliould  be  llic  liciglit  of  the  wall  in  order  that  all  the 
joints  above  the  base  may  be  frictionally  stable?  Coefficient  of  friction 
=  unity.  Ans.    12  ft. 

14.  A  wall  12  ft.  high,  2  ft.  wide  at  the  top,  and  3  ft.  wide  at  the  bot- 
tom, is  constructed  of  masonry  weighing  120  lbs.  per  cubic  foot.  The 
overturning  force  on  the  rear  face  of  the  wall,  which  is  plumb,  is  a  hori- 
zontal force  P  acting  at  4  ft.  from  tiie  base.  Find  P  so  that  the  devia- 
tion of  the  centre  of  pressure  in  the  base  may  not  exceed  \  ft.  The 
centre  of  pressure  being  fi.xed  at  2  in.  from  the  middle  of  the  base,  show 
that  I  of  the  section  may  be  removed  without  altering  its  stability,  ami 
find  the  increase  in  the  inclination  of  the  resultant  pressure  on  the  b;ise 
to  the  vertical,  consequent  on  the  removal. 

Ans.  360  lbs.;  tangents  of  angles  are  in  ratio  of  5  to  3. 

15.  A  reservoir  wall  is  4  ft.  wide  at  top,  has  a  front  batter  of  i  in  12.  ;i. 
rear  batter  of  2  in  12,  and  is  constructed  of  masonry  weighing  125  lbs. 
per  cubic  foot ;  the  maximum  compression  is  not  to  exceed  12,800  lbs. 
per  square  foot.     Find  the  limiting  height  of  the  wall. 

Ans.  24  ft.,  q  being  %% 

16.  A  dock-wall,  plumb  at  the  rear  and  having  a  face  with  a  batter  of 
I  in  24,  is  20  ft.  high  and  9  ft.  wide  at  the  base.  Counterforts  are  built 
at  intervals  of  12  ft.,  projecting  3  ft.  from  the  rear  and  6  ft.  wide. 
Determine  the  thickness  of  an  equally  strong  wall  without  counterforts, 
with  the  same  face-batter  and  also  plumb  in  the  rear. 

Ans.  10.95  ^'^• 

17.  If  the  walls  in  the  preceding  question  are  founded  in  earth  weigh- 
ing 112  lbs.  per  square  foot  and  having  an  angle  of  repose  of  32', 
find  the  least  depth  of  foundation  in  each  case,  the  masonry  weighing 
125  lbs.  per  cubic  foot.  Ans.  2.72  ft.  ;  2.71  ft. 

18.  A  vertical  retaining-wall  is  strengthened  by  means  of  vertical 
rectangular  anchor-plates  having  their  upper  and  lower  edges  18  and  22 
ft.,  respectively,  below  the  surface.  Find  the  holding  power  per  foot  of 
width,  the  earth  weighing  130  lbs.  per  cubic  foot  and  having  an  angle  of 
repose  of  30°.  Ans.  27,733^  lbs. 

19.  Determine  the  limiting  depths  of  foundation  for  {a)  a  wall  of 
rectangular  section  20  ft.  high  ;  (U)  for  a  wall  of  trapezoidal  section  hav- 
ing plumb  rear  and  front  faces  4  and  20  ft.  high  respectively.  Angle 
of  '■epose  of  earth  =30°;  weight  of  earth  =  112  lbs.  per  cubic  foot; 
of  masonry  =  140  lbs.  Ans.  (a)  3.22  ft.;  ((5)  1.93  ft. 

20.  A  wall  20  ft.  high  and  6  ft.  thick  retains  earth  on  one  side  level 
with  the  top,  and  on  the  other  the  earth  rises  up  the  wall  at  its  natural 

iope,  viz.,  45°,  to  the  height  of  5  ft.     Will  the  wall  stand  or  fall  ? 
(Weight  of  masonry  per  cubic  foot  =  130  lbs.;  of  earth  =  120  lbs.) 
Fmd  the  locus  of  the  centres  of  pressure  of  successive  layers. 


m 


EXAMPLES. 


297 


Ans.  Overturning  moment  =  4128  ft. -lbs  ;  moment  of  stability 
=  93600^  +  750 (Y  —  (yq)  =  3691 2j  ft.-lbs    \[q  =.\, 
The  wall  is  stable. 

21.  The  upper  half  of  the  section  of  a  masonry  wall  is  a  rectangle 
4  ft.  wide,  and  the  lower  half  a  rectangle  6  ft.  wide,  one  face  being  plumb. 
Find  the  height  of  the  wall  so  that  the  stress  on  the  base  may  nowhere 
exceed  10,000  lbs.  per  square  foot  when  the  wall  retams  water  (a)  on  the 
plumb  face,  (/')  on  the  stepped  face. 

(Masonry  weighs  125  lbs.  per  cubic  foot.) 

Ans.   {a)  13.08  ft. ;  {h)  9.8  ft. 

22.  A  masonry  dam  h  ft.  high  is  a  right-angled  triangle  ABC  in  sec- 
tion, and  retains  water  on  the  vertical  (AceA/J.    Show  that  the  thickness 

i  of  the  base  BC  is  given  bv  (^  —  — -,  at  bemg  the  deviation  of  the 

5(6^^  +  1) 

centre  of  pressure  in  the  base  from  the  middle  point. 
Also  show  that  the  thickness  will  be  given  by  /"  = 


^Ji^ 


if  the 

3(6$'  + I) 

rock  upon  which  the  wall  is  built  is  seamy,  and  if  it  is  assumed  that  the 
communication  between  the  water  in  the  seams,  and  that  in  the  reservoir 
produces  an  upward  pressure  upon  the  base  BC,  varying  uniformly  from 
that  equivalent  to  the  head  at  B  to  nil  at  C.  If  ^  =  \,  show  that,  in 
order  that  the  wall  may  slide,  the  coefficient  of  friction  must  be  less  than 
67  per  cent  in  the  first  and  81  per  cent  in  the  second  case. 

(Weight  of  a  cubic  foot  of  masonry  =  2|  x  weight  of  cubic  foot  of 
water.) 

23.  A  wall  30  ft.  high  is  of  triangular  section  ABC,  the  face  AB  being 
plumb,  and  water  being  retained  on  the  side  AC  level  with  the  top  of  the 
wall;  the  masonry  weighs  125  lbs.  per  cubic  foot.  Find  the  thickness  of 
the  base.5C  (a)  when  q  =^\\  (p)  when  stress  in  masonry  is  not  to  exceed 
10,000  lbs.  per  square  foot ;  {c)  when  q  =^\  and  the  wall  also  retains  earth 
on  the  side  AB  level  with  the  top,  the  angle  of  repose  being  30°. 

Ans.  (ii)  17.69  ft.;  {b)  13.19  ft.;  {c)  17  ft. 

24.  A  wall  4  ft.  wide  at  the  top,  with  a  front  batter  of  i  in  8,  and  a 
rear  batter  of  i  in  12,  is  30  ft.  high.  Will  the  wall  be  stable  or  unstable 
(i)  when  it  retains  water  level  with  the  top  ;  (2)  when  it  retains  earth  ? 

(Weight  of  masonry  per  cubic  foot  =  125  lbs. ;  of  earth  =112  lbs. ; 
angle  of  repose  =  30°  ,-  and  q  —  f .) 

Ans.  (i)  Moment  of  wt.  =  128,863  ft.-lbs.  ;  overturning  moment 
=  281,250  ft.-lbs.,  artd  wall  is  therefore  unstable. 
(2)  Moment  of  wt.  =  148,251   lbs.;   overturning  moment 
=  168,000  lbs.,  and  wall  is  therefore  unstable. 

25.  The  faces  of  a  reservoir  wall  4  ft.  wide  at  top  and  40  ft.  high  have 
the  same  batter,  and  water  rises  on  one  side  to  within  6  ft.  of  the  top. 
Find  the  batter,  assuming  {a)  that  tb«  pressure  on  the  horizontal  base  is 


B!i 


I 


298 


THEORY  OF  STRUCTURES. 


to  be  nowhere  negative ;  {b)  that  the  pressure  varies  uniformly  and  at 
no  point  exceeds  10,000  lbs.  per  square  foot. 

(Weight  of  masonry  =  125  lbs.  per  cubic  foot.) 

Ans.  (a)  35.8  ft.;  (d)  30  ft. 

26.  The  faces /i  A -^C  of  a  wall  are  parabolas  of  equal  parameters  hav- 
ing their  vertices  at  B  and  C;  water  rises  on  one  side  to  the  top  of  the 
wall.  Determine  the  thickness  of  the  horizontal  base  BC,  {a)  for  a  wail 
50  ft.  high;  (6)  for  a  wall  100  ft.  high,  so  that  the  pressure  on  the  base 
may  at  no  point  exceed  10,000  lbs.  per  square  foot.  Also  (c)  compare 
the  volume  of  such  wall  with  the  volume  of  an  equally  strong  wall  of  the 
same  height,  but  with  a  section  in  the  form  of  an  isosceles  triangle  with 
its  vertex  at  A. 

(Weight  of  masonry  =  125  lbs.  per  cubic  foot.) 

Ans.  (rt)  32.44  ft.;  (d)  119.17  ft.    

(c)  in  case  (a)  ratio  =  7  :  |/i  18 ; 
(d)     •'     =  ^Ts6  :  21. 

27.  The  water-face  AC  of  a  wall  has  a  batter  of  i  in  10 ;  the  width  of 
the  wall  AD  at  the  top  is  6  ft. ;  the  rear  of  the  wail  DEF  has  two  slopes, 
DE,  having  a  batter  of  2  in  10,  and  EF,  a  batter  of  78  in  100;  the  masonry 
weighs  125  lbs.  per  cubic  foot,  and  the  maximum  compression  must  not 
exceed  85  lbs.  per  square  inch.  Find  the  safe  heights  of  the  two  portions 
AE  and  EC. 

28.  The  section  ABCD  of  a  retaining-wall  for  a  reservoir  has  a  verti- 
cal face  ^Cand  a  parabolic  water-face  AD,  with  the  vertex  at  D.  The 
width  of  the  base  DC  =  4  x  width  of  the  top  AB.  If  AB  =  6  ft.,  find 
the  height  of  the  wall,  and  trace  the  curves  of  resistance  {a)  when  the 
reservoir  is  full ;  (i)  when  empty. 

(Cubic  foot  of  masonry  =  2  x  cubic  foot  of  water.) 
Ans.  32  ft.  if  ^  =  ^,  and  then  max.  compn.  =  8000  lbs.  per  sq.  ft. 

29.  The  figure  represents  the  section  of  the  upper  portion  of  a  masonry 
dam  which  has  to  retain  water  level  with  the  top  of 
the  dam.  The  face  AC  is  plumb  for  a  depth  of  7}  ft. 
The  width  of  the  section  is  constant  and  =  22^  ft.  for 
a  depth  AB  =  40  ft. 

F  Find  the  maximum  stress  in   the  masonry  at  the 

\  horizontal  bed  BF.     With  the  same  maximum  stress, 

\  what  should  be  the  width  of  the  horizontal  bed  CO, 

\        FG  being  straight  ? 
G  (Masonry  weighs  130  lbs.  per  cubic  foot.) 

^■°-  '^T-  '  Ans.  20,720  lbs.  per  sq.  ft. 

30.  A  wall  of  an  isosceles  triangular  section  with  a  base  36  ft.  wide 
has  to  retain  water  level  with  its  top.  How  high  may  such  a  wail  be 
built  consistent  with  the  condition  that  the  stress  in  the  masonry  is 
nowhere  to  exceed  10.546I  lbs.  per  square  foot  ? 

(Weight  of  masonry  per  cubic  foot  =  125  lbs.) 

Ans.   54  ft.,  and  q  =  j'ff. 


B 


till,; 


n 


EXAMPLES. 


299 


31.  When  a  cylindrical  bar  is  twisted,  show  that  it  is  subjected  to 
shears  along  transverse  and  radial  longitudinal  sections,  or  to  tensions 
and  compressions  on  helices  at  45°  to  the  axis. 

32.  Find  the  work  done  in  gradually  and  uniformly  compressing  a 
body  of  volume  V\  to  the  volume  Fa,  p  being  the  final  intensity  of 
pressure  and  k  the  modulus  of  compression.  Also  show  that  the 
intensity  of  stress  is  constant  throughout  the  body. 

Ans.    t^. 

33.  A  bar  is  stretched  under  a  force  of  intensity/.  If  the  bar  is  pre- 
vented from  contracting,  find  the  lateral  stress;  also  find  the  extension. 

P         P  '"^ 


Ans. 


j)t  —  2 


m  —  I       iL    m  {in  —  i ) 

34.  Taking  the  value  of  the  coefficient  of  elasticity  {E)  and  the  co- 
eiiicient  of  rigidity  (C)  to  be  15,000  and  5750  tons  for  steel,  13,950  and 
5450  tons  for  vvrouglit-iron,  and  9500  and  3750  tons  for  cast-iron,  find  the 
coeflicient  of  elasticity  of  volume  (A'), and  also  the  values  of  the  direct 
elasiicity  {A)  and  the  lateral  elasticity  (A),  assuming  the  metals  to  be 
isotropic. 

m  K  A 

3f  i2777i  ^G 

3ll        io559f  W^ 

3f  6785^  ^G 

35.  A  body  is  distorted  without  compression  or  expansion ;  find 
the  work  done. 


Ans,  Steel 

Wrought-iron. , 
Cast-iron 


A. 
^G 
WG 

%G 


Ans. 


i,f^ 


Nx'  +  AV  +  N»'  +  2(T,'  +  Ti'  +  T,')ldS. 


36.  Find  the  work  required  to  twist  a  hollow  cylinder  of  external 
radius  /?i,  internal  radius  /?a,  and  length  /through  an  angle  a, 

Ans.   ;u!^V.*  -  ^a*). 

41 

Prove  that  torsion  is  equivalent  to  a  shear  at  each  point. 

37.  Show  thai  a  simple  elongation  is  equivalent  to  a  cubical  dilation 
and  a  pair  of  shearing  or  distorting  stresses. 

38.  Find  the  resultant  shearing  stress  at  any  point  in  the  surface  of 
the  transverse  section  of  an  elliptic  cylinder.     (Art.  24,  Case  d.) 

nG     ^V 


Ans.  2O- 


i,/ being  the  perpendicular  from  the  centre 


P6'  +  c' 

upon  the  tangent  to  the  ellipse  at  the  given  point,  and 
2d,  2c  the  major  and  minor  axes. 
39.  A  cylinder  undergoes  torsion  round  its  axis.      Show  that  the 
curves  of  no  traction  are  concentric  circles. 


ai...: 

■  1  , 

( 


: 


,  i    V. 


CHAPTER   V. 
FRICTION. 

I.  Sliding  Friction. — Friction  is  the  resistance  to  motion 
which  is  always  developed  when  two  substances,  whether  solid, 
liquid,  or  gaseous,  are  pressed  together  and  are  compelled  to 
move  the  one  over  the  other.  If  /*  is  the  mutual  pressure,  and 
if  /'"  is  the  force  which  must  act  tangentially  at  the  point  of 
•contact  to  produce  motion,  the  ratio  oi  F  to  P  is  called  the  co- 
efficient of  friction  and  "may  be  denoted  by/".  The  value  of/ 
does  not  depend  upon  the  nature  of  any  single  substance,  but 
upon  the  nature  and  condition  of  the  surfaces  of  contact  of  a 
pair  of  substances.  It  is  not  the  same,  e.g.,  for  iron  upon  iron 
as  for  iron  upon  bronze  or  upon  wood  ;  neither  is  it  the  same 
when  the  surfaces  are  dry  as  when  lubricated. 

The  laws  of  friction  as  enunciated  by  Coulomb  are  : 

(i)  That  /  is  independent  of  the  velocity  of  rubbing ;  (2) 
that /is  independent  of  the  extent  of  surface  in  contact;  (3) 
that /"depends  only  on  the  nature  of  the  surfaces  in  contact. 

The  friction  between  two  surfaces  at  rest  is  greater  than 
when  they  are  in  motion,  but  a  slight  vibration  is  often  suffi- 
cient to  change  the  friction  of  rest  to  that  of  motion. 

Morin's  elaborate  friction  experiments  completely  verified 
these  laws  within  certain  limits  of  pressure  (from  |  lb.  to  128 
lbs.  per  square  inch)  and  velocity  (the  maximum  velocity  being 
10  ft.  per  second),  and  under  the  conditions  in  which  they 
were  made. 

A  few  of  his  more  important  results  are  given  in  the  follow- 
ing table : 

300 


SLIDING  FRICTION. 


3OT 


Material. 


Wood  on  wood 

Metal  on  wood 
i<      II        ■• 

Metal  on  metal 

Mclal  and  wood 

on   each    other 

or  each  on  itself 


State  of  Surfaces. 

dry 

dry 

wet 

dry 

wet 

slightly  oily 

occasionully  lubricated  as  usual 
constantly  lubricated 


Coelticient  of  Kriciiun. 


5 
6 
26 
2 


.35  to 

.2       " 

22    " 

•15    " 

■3 
•15 
.07  to  .08 

.05 


Tlic  appar.itus  employed  in  carrying  out  these  experiments 
consisted  of  a  box  whicli  could  be  loaded  at  pleasure,  and 
which  was  made  to  slide  along  a  horizontal  bed  by  means  of  a 
cord  passing  over  a  pulley  and  carrying  a  weight  at  the  end. 
The  contact-surfaces  of  the  bed  and  bo.x  were  formed  of 
the  materials  to  be  experimented  upon.  The  pull  was  meas- 
ured and  recorded  by  a  spring  dynainometer. 

More  recent  experiments,  however,  have  shown  that 
Coulomb's  laws  cannot  be  regarded  as  universally  applicable, 
but  that  /  depends  upon  the  velocity,  the  pressure,  and  the 
temperature.  At  very  low  velocities  Morin's  results  have 
been  verified  (Fleeming  Jenkin).  At  high  velocities  /  rap- 
idly diminishes  as  the  velocity  increases.  Franke,  having 
carefully  examined  the  results  of  various  series  of  experi- 
ments,  especially  those  of  Poiree,  Bochet,  and  Galton,  has 
suggested  the  formula 

/  =  /„-«^ 

V  being  the  velocity  and/,,  a,  coefficients  depending  upon  the 
nature  and  condition  of  the  rubbing  surfaces. 

For  example, 

/,  =  .29  and  a  =  .04  for  cast-iron  on  steel  with  dry  sur- 
faces. 

/„  =  .29  and  a  =  .02  for  wrought-iron  on  wrought-iron  with 
dry  surfaces.      .;      .  » 

/„  =  .24  and  a  =  .0285  for  wrought-iron  on  wrought-iron 
with  slightly  damp  surfaces. 

Ball  has  shown  that  at  very  low  pressures  /  increases  as 


a 


30« 


THEORY  OF  STRUCTURES, 


the  pressure  diminishes,  while  Rcnnie's  experiments  indicate 
that  at  very  high  pressures  y  rapidly  increases  with  the  press- 
ure,  and  this  is  perhaps  partly  due  to  a  depression,  or  to  an 
abrasion  of  the  rubbing  surfaces. 

2.  Inclined  Plane. — Let  a  body  of  weight  P  slide  uni- 
formly up  an  inclined  plane  under  a  force  Q  inclined  at  an 
angle  (i  to  the  plane. 

Let /^  be  the  friction  resisting  the  mo. 
tion,  R  the  pressure  on  the  plane,  and  a  the 
plane's  inclination. 

The  two  equations  of  equilibrium  are 

F  ■=  Q  cos  /?  —  /*  sin  « 
and 

R—  —  Q  sin  /3-\-P  cos  a. 

Qcosfi  —  Psina  „.  ,,.    . 

=  coefficient  of  friction  =/. 


K 


—  Q  sin  fi  -}-  P  cos  a 


I .  ■? 


:^1 


Let  the  resultant  of  F  and  R  make  an  angle  (p  with  the 
normal  to  the  plane.     Then 


n 

I  si 


F         Q  cos  ft  —  P?,in  n 
tan  0  =  -5-  = 


R~  -  Qs\r\  ft^Pcosa' 


or     -f;  = 


Q_  sin  {a  -f  0) 


P       cos  (ft  —  (p)' 


(/)  is  called  the  afi_§-/e  of  friction.  It  has  also  been  called  the 
angle  of  repose,  since  a  body  will  rem..; in  at  rest  on  an  inclined 
plane  so  long  as  its  inclination  does  rot  exceed  the  angle  of 
friction. 

If  a  =  o  =  /?,     then    ^  =  tan  0  =  /. 


The  work  done  in  traversing  a  distance  x  =■  Q  cos  /S.jtr.    If 

Q  is  variable,  the  work  done  —.    I    Q  cos  ft .  dx. 

3.  Wedge. — The  wedge,  or  key,  is  often  employed  to  con- 
nect members  of  a  structure,  and  is  generally  driven  into  posi- 


WEDGE. 


303 


tion  by  the  blow  of  a  hammer.     It  is  also  employed  to  force 
out  moisture  from  materials  by  induc- 
'"o  '^  pressure  thereon. 

The  figure  represents  a  wedge  de- 
scending vertically  under  a  continuous 
pressure  P,  thus  producing  a  lateral 
motion  in  the  horizontal  member  C, 
which  must  therefore  exert  a  pressure 
0  upon  the  vertical  face  AB. 

The  member  //  is  fixed,  and  it  is 
assumed  that  the  motion  of  the  machine 
is  uniform,  so  that  the  wedge  and  67  are 
in  a  state  of  relative  equilibrium. 

Let  K^ ,  R,  be  the  reactions  at  the  faces  DE,  DF,  respec- 
tively, their  directions  making  an  angle  0,  equal  to  the  angle 
of  friction,  with  the  normals  to  the  corresponding  faces. 

Let  a  be  the  angle  between  DE  and  the  vertical,  a'  the 
angle  between  DF  and  the  vertical. 

Consider  the  wedge,  and  neglect  its  weight,  which  is  usually 
inappreciable  as  compared  with  P. 

Resolving  vertically, 

R,  cos  (90°  -  a  +  0)  +  /?,  cos  (90°  -  «'-f-0) 

=  P=7?,sin(a-f  0)  +  /?,  sin(a'-;-0).    .     (i) 

Resolving  horizontally, 

R,  sin  (90"  —  a  +  0)  —  R^  sin  (90°  -  a' +  0)  =  o, 
or 

R^  cos  (or  -|-  0)  =  R^  cos  (a'  -|-  0) (2) 

Consider  the  member  C,  and  neglect  its  weight. 
Resolving  horizontally, 

R,  cos  («  -f  0)  =  e  =  ^,  cos  {a'  +  0) (3) 

Assuming  the  wedge  isosceles,  as  is  usually  the  case,  a  =  a', 
and  hence, 

by  eq.  (2),  R,  =  R^,  and  by  eq.  (i),  2R,  sin  (a  -|-  0)  =  P.       (4) 


)  I-, 


•.  ii 


I.     Ij.   i.tl 


^ 


3CM  nu-.oKv  01-  s7'A'rc/'CA'/i.s. 

Hence,  by  cqs.  (^^)  ami  (4), 

Q      cot  (fir -f  0)  _  cxtcnial  icslstancc  overcome 

(A\A\ — This  ralii)  <if  rcsisliUicc  to  riToit  is  Icrnicd  llu;  tinr/iiin 
u<t/  ttffvotffifjt^'^t-,  \n/*nt(/f,fsr,  of  a  lUiiLliiiio.) 

8u|)posc  tl\c  motion  of  the  macliiiiG  icvci'scd,  no  that  (J  \)v- 
conics  the  elToit  ami  /'  llu-  resistance. 

The  reactions  A', ,  A',  now  fall  /it-Zotc  the  notnials,  and  llu 
c«]uations  of  relative  eiiiiilil)iiiiin  art;  the  s.inic  .is  the  ahove, 
with     -  V  substituted  for  0. 


Thus, 


(J 


p  =  i  cot  (tt  -  0). 


(r.) 


The  two  cases  niay  be  included  in  the  expression 


y    =    i  cot    {(V    ±    0) (7^ 


For  a  pt'vcn  value  of  /'.  (J  incrcaj^cs  with  <v, 
If  there  were  no  friction,  0  wonkl  be  zero,  aiul  eq.  (7)  wouL 
become 


Q 
P 


cot  ft 

3 


Thus,  the  efTect  of  friction  may  be  allowed  for.  by  assuming'; 
the  wct-ls^e  frictionless,  but  with  an  an^lo  t/tirtttsri/  by  20  in  the 
/irsr  case,  and  diniiiiisfitii  b\-  J0  in  the  scioiitf  c.ise. 

Ai^ain.  whei\  P  is  the  effort  ami  Q  the  resistance,  cq.  (5) 

Q 
shows  that  if  <»'-["•/•>  90.  the  ratio  .y  is  negative,  which  is 

impossible,  while  if  «>--}- 0  =  90°,  .y  is  zero,  and  in  order  to 


WHDGH, 


305 


ovcuomo  C.'.  Iinvvcvcr  Hiniill  il  iiiif;lil  be,  /'would  rccjuiic  tu  he 
iiifmitcly  giciit,     llciicc, 

»  |-  (/»  imiHt  be  <  yo", 

.111(1  hrlnw  this  limit  ;,  diniinlHhcH  aH  0  Incrcnscs. 

Similiirly.  il  may  be  hIiovvii  fiom  ctj.  (7)  that  when  Q  is  the 
rffnrt  .md  /'  ihr  icHistaiicc, 

0  must  be  <  or, 

Q 
and  that  bcU)W  this  limit  .,  incrcaHCH  with  0. 

Elftcicmy.  iJuiin^^  llic  unifoiiii  mution  of  the  machine,  let 
any  point  a  descend  veitieally  to  the  point  h.  The  correspond- 
ing hoii/onlal  di^piatemcrtl  in  evidently  .iln . 

The  motive  work  -  -  /'.  ah  ; 
"    useful  work         Q.ibc, 

Q'^fx:      Q 
Hence,  the  emcicncy --:    ,,     ,  —  ,,  .  2   tana 

I  .all       r 

=  tan  (V  cot  [(y   |-  (/'),  by  ecj.  (5). 

This  is  a  maximum  for  a  ^dven  value  of  «/» when 


a  =  45 


2' 


;uul    the    max.    eflficiency  --  tan  (45"  —      )  (d  (45" -f- -j 

\  _  I  —  sin  <l> 

(J>  ]~  I  -{-  sin  0" 


I  -  tan   -^  ' 

2 


I  -f  tan 

For  the  reverse  motion,  the  efficiency 
P.ah 


Q.2ln 


'  =  cot  a  tan  («  —  0). 


Fin 


« : 


|.-4 


-1 


H 


'I } 


306 


THEORY  OF  STRUCTURES. 


0 


This  is  a  maximum  when  a  ■=.  45°  -j — .     Thus  the 


max.  efficiency 


(45" +f) 


cot  ^45° +  t)  tan  (45' 


♦  0\       I  —  sin  0 
I  -\-  sin  0' 


4.  Screws. — A  screw  is  usually  designed  to  produce  a 
hnear  motion  or  to  overcome  a  resistance  in  the  direction  of  its 
length.  It  is  set  in  motion  by  means  of  a  couple  acting  in  a 
plane  perpendicular  to  its  axis.  A  reaction  is  produced  be- 
tween the  screw  and  nut  which  must  necessarily  be  equivalent 
to  the  couple  and  resistance,  the  vioiion  being  steady. 

Take  the  case  of  a  .y^?^rt:rr  * -threaded  screw.  It  may  be 
assumed  that  the  reaction  is  concentrated  along  a  helical  line, 
whose  diameter,  d,  is  a  mean  between  the  external  and  internal 
diameters  of  the  thread,  and  that  iT:s  distribution  along  this 
line  is  uniform.  It  will  also  be  supposed  that  the  axes  of  the 
couple  and  screw  are  coincident,  so  that  there  will  be  no  lateral 
pressure  on  the  nut. 

Let  M  be  the  driving  couple. 
"     Q   '*     "     axial    resistance  to   be  over- 
come. 
"      r   "     "     reaction  at  any  point  a  of  the 
helical  line,  and   let   0  be 
angle  between  its  direction 
and  the  normal  at  « ;  0  is 
the  angle  of  friction. 
"      a   "     "     angle    between    the    tangent 
at  a  and  the  horizontal  ;  a 
Fig.  340.  is  called  the  pitch-angle. 

Since  the  reaction  between  the  screw  and  nut  must  be 
equivalent  to  M  and  Q,  then 


*Squarethreaded  screws  work  more  accurately  than  those  with  a  V-thread, 
but  the  efficiency  of  the  latter  has  be --n  shown  to  be  very  little  less  than  that  of 
the  former  (Poncclet).  On  the  other  hand,  the  V-thread  is  the  stronger,  much 
less  metal  lieing  removed  in  cutting  it  than  is  the  case  with  a  square  thread. 
Again,  with  a  V-thread  there  is  a  tendency  to  burst  the  nut,  which  does  not 
obtain  in  a  screw  with  a  square  thread. 


H< 


IF- 


•WPP 


f^ 


.I^J" 


li!' 


« 


SC/HE  WS. 


307 


1;  ' 


Q  =  algebraic  sum  of  vertical  components  of  the  reac- 
tions at  all  points  of  the  line  of  contact, 


=  2lr  cos  {a  -|-  0)]  =  cos  {a  -{-  (p)2{r), 


(0 


and  M  =  algebraic  sum  of  the  moments  with  respect  to  the 
axis  of  the  horizontal  components  of  the  reactions  at  all  points 
of  the  line  of  contact, 


=  ^    r  sin  (a  +  0)  ^     =  -sin  (or  -\-  (p)2{r). 


(2) 


Let  the  couple  consist  of  two  equal  and  opposite  forces,  P, 
acting  at  the  ends  of  a  lever  of  length/,  so  that  M  =  Pp. 
Hence,  by  eqs,  (1)  and  (2), 

J7  =  7^=^^°'(^+^)' 


and  the  mechanical  advantage 


(3) 


Q      s/ 

If  0  =  o,    p-  ==  ~7  cot  a,  and  the  effect  of  friction  may 

be  allowed  for,  by  assuming  the  screw  frictionless,  but  with  a 

pitch-angle  equal  to  a  -\-  0. 

Again,  let  the  figure  represent  one  complete  turn  of  the 

thread  developed  in  the  plane  of  the  '  0 

paper.  CD  is  the  corresponding  length 

of  the  thread  ;  DE  the  circumference 

nd\   CE,  parallel  to  the  axis,  the  pitch   jj" 

h\  and  CDE  i\\Q  pitch-angle  or.  Fig.  541. 

The  motive  work  in  one  revolution         =  M.  27t  =  Pp .  2n, 
The  useful  work  done  in  one  revolution  =  Qli. 

Hence,  the  efficiency  =  7, =  ~  cot  («  -f-  0) 


h 


— -,  cot  {a  -|-  0)  =  tan  a  cot  {a  -\-  0). 


(4) 


:  I 


i     ! 


',  I 


308 


THEORY  OF  STRUCTURES. 


0   .. 


This  is  a  maximum  when  a  =  45° ,  its  value  then  being 


tan 


I  -\-  tan 


In  practice,  however,  a  is  generally  much  smaller,  efficiency 
being  sacrificed  to  secure  a  large  mechanical  advantage,  which, 
according  to  eq.  (3),  increases  as  a  diminishes. 

If  «f  -f-  0  —  90°,  -^  =  o,  so  that  to  overcome  Q,  however 

small  it  may  be,  would  require  an  infinite  effort  P. 


I 


: 


.-.    ff  -|-  0   <  90°. 

Suppose  the  pitch-angle  sufficiently  coarse  to  allow  of  the 
screw  being  reversed.  Q  now  becomes  the  effort  and  P  the 
resistance.  The  direction  of  r  falls  on  the  other  side  of  the 
normal,  and  the  relation  between  P  and  Q  is  the  same  as 
above,  —  0  being  substituted  for  0. 

Thus, 

^  =  -^  cot  (a  -  0), 
and  therefore  the  mechanical  advantage 


=  ^  =  T^t^"(^-'^)' 


P 


If  a  =  </),  —  =  o,  and  to  overcome  P,  however  small  it 
may  be,  Q  would  require  to  be  infinite. 

.-.  <)/  >  0. 

If  a  <  0,  reversal  of  motion  is  impossible,  and  the  screw 
then  possesses  the  property,  so  important  in  practice,  of  serv- 
ing to  fasten  securely  together  different  structural  parts,  or  of 
locking  machines. 


m 


11 


!0«l? 


!i!T;: 


ENDLESS  SCREWi. 


309 


Again,  it  may  be  necessary  to  take  into  account  the  friction 
between  the  nut  and  its  seat,  as  well  as  the  friction  at  the  end 
of  the  screw.  The  corresponding  moments  of  friction  with 
respect  to  the  axis  are  (Art.  8) 


/ 


id:-d: 


and 


3 


/  being  the  coefificient  of  friction,  d^ ,  d^  the  external  and  inter- 
nal diameters  of  the  seat,  and  d'  the  diameter  of  the  end  of 
the  screw. 

5.  Endless  Screws  (Fig.  242). — A  screw  is  often  made 
to  work  with  a  toothed  wheel,  as,  for  ex- 
ample, in  raising  sluice-gates,  when  the  screw 
is  also  made  sufficiently  fine  to  prevent,  by 
friction  alone,  the  gates  from  falling  back 
under  their  own  weight.  The  theory  is  very 
similar  to  the  preceding.  Let  the  screw  drive. 
A  tooth  rises  on  the  thread,  and  the  wheel 
turns  against  a  tangential  resistance  Q,  which 
is  approximately  parallel  to  the  axis  of  the 
screw. 

Let  Fig.  243  represent  one  complete  turn  of  the  thread 

developed  in  the  plane  of  the  paper,  a 
being  the  pitch-angle  as  before. 

Consider  a  tooth.  It  is  acted  upon 
by  (2  in  a  direction  parallel  to  the 
axis,  and  by  the  reaction  R  between 
the  thread  and  tooth,  making  an  angle 
0  (the  angle  of  friction)  with  the  normal 
to  the  thread  CD. 


Fig.  242. 


Fig.  343. 


.'.  Q=  R  cos  (a  +  0). 


Again,  the  horizontal  component  of  R,  viz.,  R  sin  (a  -f-  0), 

d 
has  a  moment  R  sin  («  -(-  0)  -  with  respect  to  the  axis  of  the 


3IO 


TIlkORY  OF  STRUCTURES. 


screw,  and  this  must  be  equivalent  to  the  moment  of  the  driv- 
ing-couple, viz.,  Pp  (Art.  4). 


.-.  Pp  =  R-  sin  {a  +  0). 


Thus  the  relation  between  P  and  Q  is  the  same  as  in  the  pre- 
ceding article. 

Similarly  if  the  wheel  acts  as  the  driver. 


^  =  -tan(«-0). 


6.  Rolling  i"  "cti-.-i.-  -The  friction  between  a  rolling  body 
and  the  surface  over  which  it  rolls  is  called  rolling  friction. 
Prof.  Osborne  Reynolds  has  given  the  true  explanation  of  the 
resistance  to  rolling  in  the  case  of  elastic  bodies.  The  roller 
produces  a  deformation  of  the  surfaces  in  contact,  so  that  the 
distance  rolled  over  is  greater  than  the  actual  distance  between 
the  terminal  points.  This  he  verified  by  experiment,  and  con- 
cluded that  the  resistance  to  rolling  was  due  to  the  sliding  of 
one  surface  over  the  other,  and  that  it  would  naturally  increase 
or  diminish  with  the  deformation.  In  proof  of  this  he  found, 
for  example,  that  the  resistance  to  an  iron  roller  on  india- 
rubber  is  Un  times  as  great  as  the  resistance  when  the  roller  is 
on  an  iron  surface.  Hence  the  harder  and  smoother  the  sur- 
faces, the  less  is  the  rolling  friction.  The  resistance  is  not 
sensibly  affected  by  the  use  of  lubricants,  as  the  advantage  of 
a  smaller  coefficient  of  friction  is  largely  counteracted  by  the 
increased  tendency  to  slip.  Other  experiments  are  yet  re- 
quired to  show  how  far  the  resistance  is  modified  by  the 
speed. 

Generally,  as  in  the  case  of  ordinary  roadways,  the  resist- 
ance is  chiefly  governed  by  the  amount  of  the  deformation  of 
the  surface  and  by  the  extent  to  which  its  material  is  crushed. 
Let  a  roller  of  weight  rr(Fig.  244)  be  on  the  point  of  motion 
under  the  action  of  a  horizontal  pull  R. 


ROLLING  FRICTIOA^. 


311 


The  resultant  reaction  between  the  surfaces  in  contact 
must  pass  through  the  point  of  intersection  of  R  and  W. 
Let  it  also  cut  the  surface  in  the  point  B. 

Let  d  be  the  horizontal  distance  between  B  and  W. 
"    p      "       vertical  "  "        B    "     R. 

Taking  moments  about  B, 


Rp  =  Wd, 


or 


R< — 


R  =  the  resistance  =  W-. 

P 

Coulomb  and  Morin  inferred, 
as  the  results  of  a  series  of  ex- 
periments, that  d  is  independent  of  the  load  upon  the  roller  as 
well  as  of  its  diameter,*  but  is  dependent  upon  the  nature  of 
the  surfaces  in  contact. 

*Dupuit's  experiments  led  him  to  the  conclusion  that  d  is   proportional  to 
the  square  root  of  the  diameter,  but  this  requires  further  verification. 
Let  n  be  the  coefficient  of  sliding  friction. 
The  resistance  of  the  roller  to  sliding  is  /<  IV,  and  "  rolling  "  will  be  insured 

d 
if  R  <  /<  W,  i.e.,  if  -  <  tan  0,  which  is  generally  the  case  so  long  as  the  direc- 

P 
lion  of  R  does  not  fall  below  the  centre  of  the  roller. 

Assume  that  R  is  applied  at  the  centre.     The  radius  r  may  be  substituted 

for/,  since  d'\s  very  small,  and  hence 

R  =  ivi 

r 

An  equation  of  the  same  form  applies  to  a  wheel  rolling  on  a  hard  roadway 
over  obstacles  of  small  height,  and  also  when  rolling'on  soft  ground.  In  the 
latter  case,  the  resistance  is  proportional  to  the  product  of  the  weight  upon  the 
wheel  into  the  depth  of  the  rut,  and  the  depth  for  a  small  arc  is  inversely  pro- 
portional to  the  radius. 

Experiments  on  the  tractional  resistance  to  vehicles  on  ordinary  roads  are 
few  in  number  and  incomplete,  so  that  it  is  impossible  to  draw  therefrom  any 
general  conclusion. 

From  the  experiments  carried  out  by  Easton  and  Anderson,  it  would  appear 
that  the  value  of  d  in  inches  varies  from  1.6  to  2.6  for  wagons  on  soft  ground, 
and  that  the  resistance  is  not  sc  ibiy  affected  by  the  use  of  springs.  Upon 
a  hard  road,  in  fair  condition,  the  resistance  was  found  to  be  from  i  to  J  of 
that  on  the  soft  ground,  the  average  value  of  d  being  \  mch,  and  was  very 
sensibly  diminished  by  the  use  of  springs. 


■  IIH! 

"     "'     !* 

m 

^■ti.^M^iiUIiihiwLn  ' 

^MHHPi' 

h\V 


'         '1'- 

iS^^^^myu 

;        1     ;..'■' 

1           1    L    i;  KH 

p.:-''^,  •:i^J 

312 


THEORY  OF  STRUCTURES. 


7.  Journal-friction. — Experiments  indicate  that  /  is  not 
the  same  for  curved  as  for  plane  surfaces,  and  in  the  ordinary 

cases  of  journals  turninjr  in  weii- 
lubricated  bearings  the  value  of/ is 
probably  governed  by  a  combina- 
tion of  the  laws  of  fluid  friction  and 
of  the  sliding  friction  of  solids. 

The  bearing  part  of  the  journal 

is  generally  truly  cylindrical  and  is 

terminated     by    shoulders     resting 

against   the   ends   of     the   step   in 

which  the  journal  turns. 

Consider  a  journal  in  a  semicircular  bearing  with  the  cap 

removed.     When  the  cap  is  screwed   on,  the  load  upon  the 

journal  will  be  increased  by  an  amount  approximately  equal 

to  the  tension  of  the  bolts.    Let  Pbe  the  load. 

Assume  that  the  line  of  action  of  the  load  is  vertical  and 
that  it  intersects  the  axis  of  the  shaft.  This  load  is  balanced 
by  the  reaction  at  the  surface  of  contact,  but  much  uncertainty 
exists  as  to  the  manner  in  which  this  reaction  is  distributed. 
There  are  two  extremes,  the  one  corresponding  to  a  normal 
pressure  of  constant  intensity  at  every  point  of  contact,  the 
other  to  a  normal  pressure  of  an  intensity  varying  from  a 
maximum  at  the  lowest  point  /^  to  a  minimum  at  the  edge  of 
the  bearing  B. 

Let  /  be  the  length  of  the  bearing,  and  consider  a  small 
element  AS  at  any  point  C,  the  radius  OC  {=  r)  making  an 
angle  0  with  the  vertical  OA. 

First.  Let  /  be  the  constant  normal  intensity  of  pressure. 


P  -  2{pAS  cos  ^. /)  =  pl-^ipU)  =  2plr. 

Frictional  resistance  =2{//>ASl)=  /p/2{AS)=/p/7rr=/P-. 

The  frictional   resistance   probably  approximates   to   this 
limit  when  the  journal  is  new. 


''     tl 


> 


JO  URN  A  L-FRICTION. 


313 


Second.  Let  /  =  /„  cos  ^, 
so  that  the  intensity  is  now  proportional  to  the  depth  CD  and 
varies  from  a  maximum  /„  at  A  to  nil  ^t  B.     This,  perhaps, 
represents  more  accurately  the  pressure   at    different  points 
when  the  journal  is  worn. 


'  -Is; 


.:  P  =  2(/ J5  cos  e.  I)  =  2[p,AS  cos'  d.  I) 
=  2pjr,   f-  cos'  6* .  ^6^  =  pjr\ 


2 


and  i\ie  frictional  resistance  =  2{/pJS/)  =  2/pJr  =  fP-. 

...  7r  4 

Hence,  the  frictional  resistance  lies  between  fP~    and  /"/*— . 

2  TT 

It  may  be  represented  by  fxP,  ix  being  a  coefficient  of  friction 
to  be  determined  in  each  case  by  experiment. 

The  total  uiomcnt  of  frictional  resistance  must  necessarily 
be  equal  and  opposite  to  the  moment  M  of  the  couple  twisting 
the  shaft ;  i.e., 

M  =  fxPr. 

Thus,  the  total  reaction  at  the  surface  of  contact  is  equiva- 
lent to  a  single  force  P  tangential  to  a  circle  of  radius  ^r  having 
its  centre  at  0  and  called  the  friction-circle. 

The  work  absorbed  by  axle-friction  per  revolution 

=  M.27t  =  2)xnPr. 
The  work  absorbed  by  axle-friction  per  minute 
=  2}ntPrN  —  piPv, 

N  being  the  number  of  revolutions  and  v  the  velocity  per 

minute. 


314 


THEORY  OF  STRUCTURES. 


' 


The  work  absorbed  by  frictional  resistance  produces  an 
equivalent  amount  of  heat,  which  should  be  dissipated  at  once 
in  order  to  prevent  the  journal  from  becoming  too  hot.  This 
may  be  done  by  giving  the  journal  sufificient  bearing  surface 
(an  area  equal  to  the  product  of  the  diameter  and  the  len<^th 
of  the  bearing),  and  by  the  employment  of  a  suitable  unguent. 

Suppose  that  h  units  of  heat  per  square  inch  of  bearing 
surface  {Id)  are  dissipated  per  minute. 

Let  /  inches  be  the  length  and  d  inches  the  diameter  of 
the  journal. 

hdl  =  heat-units  dissipated  =  heat-units  equivalent  to 
frictional  resistance 


12/       ~I^' 


J  being  Joule's  equivalent,  or  778  ft.-lbs. 


12/ h      PN        ,     12//1      Pv 
and 


ixn         I 


fx    ~  Id  ' 


Let  jz  =p  =  pressure  per  square  inch  of  bearing  surface. 


Id 


\2jh 

.  pv  = =  a  constant. 

M 


In  Morin's  experiments  ^^  varied  from  2  to  4  in.,  P  from 
330  lbs.  to  2  tons,  and  v  did  not  exceed  30  ft.  per  minute,"  so 
that/z'  was  <  5CX)0,  and  the  coefificient  of  friction  for  the  given 
limits  was  found  to  be  the  same  as  for  sliding  friction. 

Much  greater  values  oi pv  occur  in  modern  practice. 

Rankine  g\vesp{v  -\-  20)  =  44800  as  applicable  to  locomo- 
tives. 

Thurston  gives  ^z^  =  60000  as  applicable  to  marine  engines 
and  to  stationary  steam-engines. 

Frictional  wear  prevents  the  diminution  of  /below  a  certain 


JOURNAL  FRICTION. 


315 


limit  at  which  the  pressure  per  unit  of  bearing  surface  exceeds 
a  value/  given  by  the  formula. 


where 


P=pld=  pkd'' ; 
/ 


k  = 


d 


In  practice  /i  =  ^  for  slow-moving  journals  (e.g.,  joint-pins), 
and  varies  from  i^  to  3  for  journals  in  continuous  motion.  The 
best  practice  makes  the  length  of  the  journal  equal  to  four 
diameters  (i.e.,  k  =  4)  for  mill-shafting. 

Again,  if  the  journal  is  considered  a  beam  supported  at 
the  ends, 


CPl 


qcP 
32" 


-^, 


q  being  the  maximum  permissible  stress  per  square  inch,  and 
C  a  coefficient  depending  upon  the  method  of  support  and 
upon  the  manner  of  the  loading. 

.*.  d"^ Of.  — . 

For  a  given  value  of  /*,  d  diminishes  as  q  increases.  Also, 
it  has  been  shown  that  the  work  absorbed  by  friction  is 
directly  proportional  to  d. 

Hence,  for  both  reasons,  d  should  be  a  minimum  and  the 
shaft  should  be  made  of  the  strongest  and  most  durable 
material.  In  practice  the  pressure  per  square  inch  of  bearing 
surface  may  be  taken  at  about  2  tons  per  square  inch  for  cast- 
iron,  3J  tons  per  square  inch  for  wrought-iron,  and  6^  tons  per 
square  inch  for  cast-steel. 

It  would  appear,  however,  from  the  recent  experiments  of 
Tower  and  others,  that  the  nature  of  the  material  might  become 
of  minor  importance,  while  that  of  a  suitable  lubricant  would  be 
of  paramount  importance.  They  show  that  the  friction  of 
properly  lubricated  journals  follows  the  laws  of  fluid  friction 
much  more  closely  than  those  of  solid  friction,  and  that  the 


3l6 


THEORY   OF  STKUCrURES. 


lubrication  might  be  made  so  perfect  as  to  prevent  any  ab- 
solute contact  between  the  journal  and  its  bearing.  The 
journal  would  therefore  float  in  the  lubricant,  so  that  there 
would  be  no  metallic  friction.  The  loss  of  power  due  to  fric- 
tional  resistance,  as  well  as  the  consequent  wear  and  tear,  would 
be  very  considerably  diminished,  while  the  load  upon  the 
journal  might  be  increased  to  almost  any  extent. 

Tower's  experiments  also  indicate  that  the  friction  dimin- 
ishes as  the  temperature  rises,  a  result  which  had  already  been 
experimentally  determined  by  Him.  It  was  also  inferred  bv 
Hirn  that,  if  the  temperature  were  kept  uniform,  the  friction 
would  be  approximately  proportional  to  sU\  and  Thurston 
has  enunciated  the  law  that,  with  a  cool  bearing,  the  friction  is 
approximately  proportional  to  Vv  for  all  speeds  exceedinj^ 
100  ft.  per  minute. 

With  a  speed  of  150  ft.  per  minute  and  with  pressures  vary- 
ing  from  100  to  750  lbs.  per  square  inch,  Thurston  found  ex- 
perimentally that /"varied  inversely  as  the  square  root  of  the 
intensity  of  the  pressure.  The  same  law,  but  without  any 
limitations  as  to  speed  or  pressure,  had  been  previously  stated 
by  Hirn. 

8.  Pivots. — Pivots  are  usually  cylindrical,  with  the  circular 
edge  of  the  base  removed  and  sometimes  with  the  whole  of 
the  base  rounded.  Conical  pivots  are  employed  in  special 
machines  in  which,  e.g.,  it  is  important  to  keep  the  axis  of  the 
shaft  in  an  invariable  position.  Spherical  pivots  are  often 
used  for  shafts  subject  to  sudden  shocks  or  to  a  lateral  move- 
ment. 

{a)  Cylindrical  Pivots, — If  the  shafts  are  to  run  slowly,  the 
intensity  of  pressure  (/>)  on  the  step  should  not  be  so  great  as 
to  squeeze  out  the  lubricant.  Reuleaux  gives  the  following 
rules: 

The  maximum  value  of  p  in  lbs.  per  square  inch  should  be 
700  for  wrought-iron  on  gun-metal,  470  for  cast-iron  on  gun- 
metal,  and  1400  for  wrought-iron  on  lignum-vitae. 

For  rapidly-moving  shafts, 

d—cVPn, 


H  ;/  being  ' 
H  to  be  det( 
™  and/^th( 
Suppc 
divided  i 
rin^rs  be  h 
In  on( 
by  the  fri 


Hence  th( 


where 


and  d^,d^i 
face  in  con 

If  the  T 
work  absorl 

Again,  t 


and  the  tota 


If^.= 


o. 


Thus,  in  ; 
times  the  mo 


til 


PIVOTS. 


3<7 


II 


id  be 
gun- 


n  being  the  number  of  revolutions  per  minute,  c  a  coefficient 
to  he  determined  by  experiment  (  =  .0045), 
,111(1  /'the  load  upon  the  pivot. 

Suppose  the  surface  of  the  step  to  be 
tiiviiled  into  rings,  and  let  one  of  these 
riiv^^s  be  bounded  by  the  radii  x,  x  -(-  dx. 

In  one  revolution  the  work  absorbed 
by  the  friction  of  this  ring 


=  /t</> .  2nx .  dx .  2nx. 
Hence  the  total  v^oxV  absorbed  in  one  revolution 


Pic.  346. 


=   /    ^iipit^x^dx 


I^P^^  /jz  jj\  2  d*  —  d^ 


where 


P=pjL^d:-d:), 


and  d^ ,  d^  are  the  external  and  internal  diameters  of  the  sur- 
face in  contact. 

If  the  zvholc  of  the  surface  is  in  contact,  d^  =  O,  and  the 
work  absorbed  =  ^;A7rPd^. 

Again,  the  moment  of  friction  for  the  ring 

=  fxp.2nx  .dx.x  =  2/u7r/>x^ .  d.  ; 
and  the  total  moment 


~7^» 


d'  —  d* 
2ixnpx'dx  =  \f^7tp  — 

in  8 


12  ^  '  ^'         3    d^'  —  d^ 

uP 
If  d^  r=  o,  the  moment  —  ^-—d^. 

3 
Thus,  in  both  cases,  the  work  absorbed  by  friction  =  2n 
times  the  moment  of  friction. 


3i8  'j7//-:oA'y  o/'-  si'A'L'crc/A'/is. 

Let  D  be  the  wmn  diameter  of  the  surface  in  contact 

.        2        ' 

Let  2y  be  the  width  of  the  surface  in  contact  =  d^  —  d^. 
Then 

work  absorbed  = /<^/^fZ?+ ~-j. 

Some»-imes  shafts  have  to  run  at  high  speeds  and  to  bear 
heavy  pressures,  as,  e.g.,  in  screw-propellers  and  turbines.  In 
order  that  there  may  be  as  little  vibration  as  possible,  /  nuist 
be  as  small  as  practicable,  and  this  is  to  some  extent  insured 
by  using  a  collar-journal. 

Let  N  be  the  number  of  collars,  and  let  </, ,  <■/,  be  the  exter- 
nal and  internal  diameters  of  a  collar. 

Then  work  absorbed  by  friction  per  revolution  per  collar 

ut>7T^ ,  ,,         ,,,        „       /'  (i.^  —  r/,'  ,  -  .     . 

=  ^^~-  {d^  —  </.')  =  iM^nr  frZT'h  ~  2;rX moment  of  friction. 

According  to  Reuleaux,  the  mean  diameter  of  a  collar 


=  -  =  7^- 


i-^ 


n  being  the  number  of  revolutions  per  minute. 

Also,  the  zvidih  of  surface  in  contact  =  </,  —  d^  =  .48  V'l), 
and  the  maximum  allowable  pressure  per  square  inch 

(d)  Wear. — The  wear  at  any  point  of  the  elementary  ring 
must  necessarily  be  proportional  to  the  friction  ///•,  and  also  to 
the  amount  of  rubbing  surfAce  which  passes  over  the  point  in 
a  unit  of  time,  i.e.,  the  velocity  A.r\  A  being  the  angular  ve- 
locity of  the  shaft. 


11 


!  ''.?  U 


X-?'A'/?J. 


PIVOTS. 

Ilencc,  the  wear  at  any  point  is  proportional  to  fjipAx, 

(f)  Conical  Pivots. — As    before,  suppose  , 

the  surface  of  the  step  to  be  divided  into  a    \    _.ri._l 
number  of  elementary  rings.     Two  cases  will 
be  discussed  : 

First.  Assume  that  the  normal  intensity 
of  pressure  /  at  the  surface  of  contact  is 
constant. 

Let  X,  X  -f-  dx  be  the  distances  of  D  and 
/;",  respectively,  from  the  axis. 

The  total  moment  of  friction 


319 


-/ 


-Xj-f  — 


Ftc.  347. 


- ,  1  ,.t'. 


=  ri^pDE  .  2nx  .x^~-^   rx^dx 

t/.rj  sin   0(  t/.r, 

3  sm  a 

.r, ,  -f,  being  the  radii  of  the  top  and  bottom  sections  of  the 
stop. 

Also,  Py  the  total  load  on  the  pivot, 

=  /    pDE  sin  a .  mx  —  mp  I     xdx 


2     uP    X  * X  * 

Hence  total  moment  of  friction  =  --.—       '  " 


3  sni  a  .r,'  -  x,' 


Si'coi:^.  Assume  that  the  wear  is  of  such  a  nature  that  every 
point,  e.g.,  D,  descends  vertically  through  the  same  distance. 
Thus,  the  normal  wear  a  sin  «, 

or     up  Ax  oc  sin  a, 

or  px  a  sin  a. 

In  the  present  case  a  is  constant,  and  hence /.r  =  a  con- 
stant. 


if  I 


320  TJiKOKV   Oh   STRUCTUKES, 

Thus,  total  moment  of  friction 


Als6, 


=  /     i.^pDh  .  2nx .  X  =  -  .- —  /     xdx 
«/*«  sin  (r  i/.i, 

SHI  l\t 

/'r::     /     y>/)/:  sin  a  .  27rx 

/■'" 
=::  2n/>x    I      dx  =■  2fr/>x{x, 


■»',). 


/'/' 


Hence  total  moment  of  friction  =  - -; — (.r,   f-.tO. 

2  SM)  tr^   '    '      '' 

(</)  Sc/iii/i's  Pivots. — Tin:  object  aimcil  at  in  these  pivots 
I  is  to  i;ive  the  step  such  a  form  that  the  wen 

and  the  pressure  are  the  same  at  all  points. 
Let  0  be  the  any;le  jnade  In-  tin-  tangent  at 
•V"-/d     any  point  of  the  step  with  the  axis. 

Let   I'  be  the  distance  of  the  point  from 
the  axis.     'Ihen 

Pj>  a  sin  6\ 
and  hence  if/>  is  constant, 
y  <x  s\n  6     or    y  cosec  ti  =  ^  const. 

is  the  equation  of  the  generating  line  of  the  step.  This  line  is 
known  as  the  tituirix  and  also  as  the  anti-friction  curve.  If 
the  tangent  at  D  intersects  the  axis  in  T, 

DT  —  y  cosec  ^  =  a  const. 

The  curve  may  be  traced  by  passing  from  one  point  to  an- 
other and  keeping  the  tangent  DT  of  constant  length. 
The  above  equation  may  be  written 

ds 

y-jr  —  a  const.  —  a, 


or 


BELTS  AND   ROVES. 


321 


or 


ds  _a  dy  _      /        tdy  \« 
~di~'y  dx  ~  Y  '  ^^  Kfxi  ' 

which  may  be  easily  integrated,  the  result  being  the  analytical 
equation  to  the  curve,  viz., 

{a  -  Vir^'f\    ,     ^-- , 

X  —  a  logJ ^/  +  ^(^  —  y  -\-ty.  const. 


Sihiele  or  anti-friction  pivots  arc  suitable  for  high  speeds,  but 
have  not  been  very  j.;enerally  adopted. 

9.  Belts  and  Ropes. — Let  the  figure  represent  a  pulley 
movable  about  a  journal  at  O,  and  let  a  belt  (or  roi)e),  acted 
upon  by  forces  1\  ,  7",  at  the  ends,  embrace  a  portion  ABC 
of  the  circumference  subtending  an  angle  ii  at  the  centre. 

in  order  that  there  may  be  motion  in  the  direction  of  the 
arrow,  T^  must  exceed  7",  by  an  amount  sufficient  to  overcome 
{\\\:  frictional  rcsistaticc  along  the  arc  of  contact  and  the  resist- 
(Uitc  to  bi'itding  due  to  the  stiffness  of  the  belt. 

Consiiler  first  the  frictional  resist- 
ance, and  suppose  the  belt  to  be  ou  the 
point  of  s/i/>/>i/fi;\ 

Any  small  element  JUi'  (=  rfs)  of 
the  belt  is  acted  upon  by  a  pull  T  tan-  ^/ 
{fcntial  to  the  pulley  at  B,  a  pull  T—  dT 
tau'^^ential  to  the  pulley  at  B' ,  and  by  a 
iciction  equivalent  to  a  normal  force  Rds 
It  the  middle  point  of  /)'/)",  and  a  tan- 
gential  force,   or   frictional    resistance, 

Let  the  angle  COD  —  8,  and  the  an- 
gle BOB'  =  de.  P'<--  '49. 
Resolving  normally, 

{T-\T-  dT)  s\n^  -  Rds  =  o.      .     .     .     (i) 


% 


mif 


m  I 


I . 


H";    I 


ja^.  •  .  THEORY  OF  STRUCTURES. 

Resolving  tangentially, 

dH 


{T  —  T  —  dT)  cos  —  —  ixRds  =  o,       ...     (2) 

/i  being  the  coefficient  of  friction.  .         . 

..         ,     .    .  „       .     dO  .  .         ,       dB 

Now  ad  being   very  small,    sin  —  is   approximately    — . 

dd  ' 

cos  —  is  approximately  unity,  and   small  quantities   of   the 

second  order  may  be  disregarded. 

Hence,  eqs.  (i)  and  (2)  may  be  written 


TdB-Rds-o (3j       !] 


and 


dT  —  jxRds  =  o. 


dT 


.-.  dT  —  jxTdB,     or    -^p  —  t^dd.  .    .    . 


Integrating, 


log,T=M(^+C, 


C  being  a  constant  of  integration. 

When  (9  =  0,     T  =  T^,     and  hence     log,  7",  =  (7. 


(4' 
(5> 


I  1 


.-.   log,-:™  =  fxO, 

or        -;f  =  ^^  • 


(6) 


When  e  =  cr,     T  -  T„    and  hence 


(7) 


t-  being   the  number  2.71828,  i.e.,  the  base  of   the  Naperian 
system  of  logarithms. 


.    (6) 


.    (7) 


iperi 


;in 


BELTS  AND  ROPES. 


323 


Fig.  250. 


If  «  is  increased  by  /?,  the  new  ratio  of  tensions  will  be 
c^^  times  the  old  ratio ;  so  that  if  a  increases  in  arithmetical 
progression,  the  ratio  of  tensions  will  increase  in  geometrical 
progression.  This  rapid  increase  in  the  ratio  of  the  tensions, 
corresponding  to  a  comparatively  small  increase  in  the  arc  of 
contact,  is  utilized  in  "  brakes" 
for  the  purpose  of  absorbing 
surplus  energy.     For  example  : 

A  flexible  brake  consisting 
of  an  iron  .or  steel  strap,  or, 
again,  of  a  chain,  or  of  a  series 
of  iron  bars  faced  with  wood 
and  jointed  together,  embraces 
about  three-fourths  of  the  cir- 
cumference of  an  iron  or  wooden 
drum.  One  end  of  the  brake 
is  secured  to  a  fixed  point  0  and  the  other  to  the  end  B  o(  a. 
lever  A  OB  turning  about  a  fulcrum  at  O.  A  force  applied  at 
.i  will  cause  the  brake  to  clasp  the  drum  and  so  produce  fric- 
tion which  will  gradually  bring  the  drum  to  rest. 

Let  a)  be  the  angular  velocity  of  the  drum  before  the  brake 
is  applied. 

Let  /be  the  moment  of  inertia  of  the  drum  with  respect  to 

its  axis. 

/a)* 
The  kinetic  energy  of  the  drum  = . 

When  the  brake  is  applied,  the  motion  being  in  the  direc- 
tion of  the  arrow,  let  the  greater  and  less  tensions  at  its  ends 
be  r, ,  T^,  respectively. 

Let  n  be  the  number  of  revolutions  in  which  the  drum  is 
brought  to  rest.     Then 

i/oj'  =  {T,  -  T,)ndn, (8) 

./being  the  diameter  of  the  drum. 

Also,  if  Pis  the  force  applied  at  A,  and  if  p  and  q  are  the 
perpendicular  distances  of  O  from  the  directions  of  P  and  7",, 

rcsp(,'ctively. 


Pp  -  T^f' 


(9) 


;:(  'I'l  Ss'j'tl 


II' 


i  ■ 


n 
I  i 

i  !■ 


m 


IH'lp 


?ii 


'     A 

■:  1 

u 

, 

.,1 

'   ! 

;          ! 

.' 

■ 

1 1' 


THEORY  OF  STRUCTURES. 


Again, 


y,  =  T^c^", (lo) 


a  being  the  angle  subtended  at  the  centre  by  the  arc  of  contact. 
Hence,  by  eqs.  (8),  (9),  (lO),   , 


n  = 


qJoa^ 


2Pp{c>"'—  \)nd' 


(n) 


If  the  motion  of  the  drum  were  in  the  opposite  direction,  q 
would  be  the  perpendicular  distance  of  0  from  the  direction  of 
Z, ,  and  then  Pp  =  T^q. 

Proceeding  as  before,  , 


n  = 


^/coV"" 


2Pp{e*"-—  \)niV 


and  therefore  the  number  of  turns  '.1  the  second  case,  before 
the  drum  comes  to  rest,  is  e^*^  times  the  number  in  the  first, 
which  is  consequently  the  preferable  arrangement. 

The  coefficient  of  friction  fi  varies  from  .12  for  greasy  shop 
belts  on  iron  pulleys  to  .5  for  new  belts  and  hempen  ropes  on 
wooden  drums.  In  ordinary  practice,  an  average  value  of  /< 
for  dry  belts  on  iron  pulleys  is  .28,  and  for  wire  ropes  .24;  if 
the  belts  are  wet,  }x  is  about  .38. 

Formulae  (6)  and  (7)  are  also  true  for  non-circular  pulleys. 

10.  Effective  Tension. — The  pull  available  for  the  trans- 
mission of  power  =  r,  —  7",  =  6".  Let  HP  be  the  horse- 
power transmitted,  v  the  speed  of  transmission  in  feet  per  sec- 
ond, a  the  sectional  area  of  the  rope  or  belt,  and  s  the  stress 
per  square  inch  in  the  advancing  portion  of  the  belt. 

ThcK,  if  T',  and  7",  are  in  pounds, 


HP 


(7-, 


-  T,)v       Sv 

=  — ,     and 

550  550' 


as. 


The  working  tensile  stress  per  square  inch  usually  adopted 
for  leather  belts  varies  from  285  lbs.  (Morin)  to  355  lbs.  (Claudcl), 


EFFECT  OF  HIGH  SPEED. 


325 


"■"Wli! 

\m 

liiMl 

* 

i 

■■■I 

f 

m 

RfHpRr 

'   i  ^  \ 

T- 

1 
^    r 

an  average  value  being  300  lbs.  In  wire  ropes,  8500  lbs.  per 
square  inch  may  be  considered  an  average  working  tension. 

Hempen  ropes  fpr  the  transmission  of  power  generally  vary 
from  4^  to  6^  in.  in  circumference. 

II.  Effect  of  High  Speed. — When  the  speed  of  trans- 
mission is  great,  tlie  effect  of  centrifugal  force  must  be  taken 
into  account. 

wads  v' 
The  centrifugal  force  or  the  element  ds  = ,  zu  being 

the  specific  weight  of  the  belt  or  rope,  and  r  the  radius  of  the 
pulley. 

Eq.  (3)  above  now  becomes 

Tdd  —  Rds =  o, 


or 


^  ,„      wadd  ,        „  , 

Tdd —~v*  —  Rds  =  o\ 


and  hence,  by  eq.  (4), 


dT 


T—  — v' 
S 


=  ^de. 


Integrating, 


T- 


wa 


log* 


g 


■v^ 


=  txe 


g 

since  T  —  T^  when  Q  —  o. 

Also,  T—T^  when  6^  =  a,  and  therefore 

wa  . 


T. 


g 


-V 


T, 


wa 
~g 


z=ei"', 


-v^ 


or 


wa 


7;  =  7;^« ■y'(>"- 1). 


1 


THEORY  OF  STRUCTURES, 

The  work  transmitted  per  second 

=  (r.-r>=:(r,z;-'^z.')(.^--i), 


which    is    a    maximum    and    equal    to    f  7',(r'*»  —  i)    when 
and  the  two  tensions  are  then  in  the  ratio  of 


V 


=   A  / ,  a 


2^^"  +  I  to  3. 

The  speed  for  which  no  work  is  transmitted,  i.e.,  the  Hniit- 
ing  speed,  is  given  by 


T.v -v' 


o,      or 


V    zva 


12.  Slip  of  Belts.— A  length  /  of  the  belt  (or  rope)  becomes 

/(i  +  -^jon  the  advancing  side  and  /(i  -\-  —j  on  the  s/acJt  side, 

T  T^ 

where  />,  =  -'  and  />,  =  --,  B  being  the  coefficient  of  elasticity. 

Thus,  the  advancing  pulley  draws  on  a  greater  length  than  is 
given  off  to  the  driven  pulley,  and  its  speed  must  therefore 
exceed  that  of  the  latter  by  an  amount  given  by  the  equation 


;  i 


4    1 


reduction  of  speed,  or  slip 
speed  of  driving  pulley 


A -A 


The  slip  or  creep  of  the  belt  measures  the  loss  of  work. 
In  ordinary  practice  the  loss  with  leather  belting  does  not  ex- 
ceed 2  per  cent,  while  with  wire  ropes  it  is  so  small  that  it  may 
be  disregarded. 


PKON  Y'S  D  YNAMOME  TEli. 


327 


13.  Prony's  Dynamometer. — This  dynamometer  is  one  of 
tiie  commonest  forms  of  friction-brake.  The  motor  whose 
power  is  to  be  measured  turns  a  wheel  E  which  revolves  be- 
tween the  wood  block  B  and  a  band  of  wood  blocks  A.     To 


A^? 


Fig.  351. 


the  lower  block  is  attached  a  lever  of  radius  p  carrying  a 
weight  P  at  the  free  end.  By  means  of  the  screws  C,  D  the 
blocks  may  be  tightened  around  the  circumference  until  the 
unknown  moment  of  frictional  resistance  FR  is  equal  to  the 
known  moment  Pp. 

The  weight  /*,  which  rests  upon  the  ground  when  the 
screws  are  slack,  is  now  just  balanced. 

The  work  absorbed  by  friction  per  minute  =  znRFn  =  mPpn, 

II  being  the  number  of  revolutions  per  minute. 

14.  Stiffness  of  Belts  and  Ropes. — The  belt  on  reaching 
the  pulley  is  bent  to  the  curvature  of  the  periphery,  and  is 
straightened  again  when  it  leaves  the  pulley.  Thus,  an  amount 
of  work,  increasing  with  the  stiffness  of  the  belt,  must  be  ex- 
pended to  overcome  the  resistance  to  bending.  As  the  result 
of  experiment,  this  resistance  has  been  expressed  in  the  form 

i>R' 


T  being  the  tension  of  the  belt,  a  its  sectional  area,  R  the 


radius  of  the  pulley,  and  6  a  coefificient  to  be  determined. 
According  to  Redtenbacher,  /f  =  2.36  in.  for  hempen  ropes. 

t(  n  a  ^  =   1.67    "       "  "  " 

«  ••  Reuleaux,  b  =  3.4     "     "    leather  belts. 


1 

r. 

!'|! 

1  1 1 

J   .  ( 

h 


,.;'!..    -'in 


lif 


f  1 


328 


THEORY  OF  STRUCTURES. 


Let  the  figure  represent  a  sheave  in  a  pulley-block  turning 

in  the  direction  of  the  arrow  about  a 
journal  of  radius  r. 

Let   T,  be  the  effort,  7!,  the  re- 
sistance. 

The  resistance  due  to  the  stiff- 
ness of  the  belt  may  be  allowed  for 

aT 
aTs  by  adding  -ttb  to  the  force  71.    The 

Fig.  2s».  frictional  resistance  at  the  journal- 

surface  is  Psin  0  or//*,  P being  the  resultant  of  7,,  7,. 
The  motion  being  steady,  taking  moments  about  the  centre, 


T,R=.\T^-\- 


Tr 


)R+fPr, 


or 


T,-T,-]r  ^^  -\rf  pP- 


If  7,  and  7,  are  parallel,  7*  =  7,  -|-  7,,  and  the  last  equa- 
tion becomes  ,' 

T.=  T,-^YR'^f'^{T,-\-T^. 

Let  the  pulley  turn  through  a  small  angle  B. 
The  counter-efficiency  of  the  sheave 


motive  work       7,#      7, 


2/r 


a 


~  useful  work  ~  TJ^~  T~^  '^  R  -  fr^  b  R  —  fr' 

In  the  case  of  an  endless  belt  connecting  a  pair  of  pulleys 
of  radius  R^,  R^,  the  resistance  due  to  stiffness  m?y  be  taken 

equal  to  -r-y-p  +  d/-  ^bemg  the  mean  tension  ^= ^ 

The  resistance  due  to  journal-friction  — /rP\j^  ~^'rJ' 
The  useful  resistance  =  T,  —  T,  =  S. 


WHEEL   AND  AXLE. 
Hence,  the  counter-efficiency 


329 


=  ^  +  (1  +  ^)11^+ "■^'■?)- 


\R,  '  Rj\bS 

In  wire  ropes  the  stress  due  to  bending  may  be  calculated 
as  follows :  * 

Let  X  be  the  radius  of  a  wire.  The  radius  of  its  axis  is 
sensibly  the  same  as  the  radius  R  of  the  pulley. 

The  outer  layers  of  the  wire  will  be  stretched,  and  the  inner 
shortened,  while  the  axis  will  remain  unchanged  in  length. 

Hence, 

X      change  of  length  of  outer  or  inner  strands      unit  stress 
R  ~  length  of  axis  ~  E         ' 

and  the  unit  stress  due  to  bending  =  E-jc . 

15.  Wheel  and  Axle. — Let  the  figure  represent  a  wheel 
of  radius  p  turning  on  an  axle  of  radius  r,  under  the  action  of 
the  two  tangential  forces  P  and  Q,  in- 
clined to  each  other  at  an  angle  d. 

The  resultant  R  oi  P  and  Q  must 
equilibrate  the  resultant  reaction  be- 
tween the  wheel  and  axle  at  the  sur- 
face of  contact. 

Let  the  directions  of  P  and  Q 
meet  in  T. 

If  there  were  no  friction,  the  re- 
sultant reaction  and  the  resultant  R 
would  necessarily  pass  through  O 
and  T. 

Taking  friction  into  account,  the 
direction  of  R  will  be  inclined  to  TO. 
Let  its  direction  intersect  the  circumference  of  the  axle  in  the 
point  A.  The  angle  between  TA  and  the  noxmaX  AO  dA  A, 
the  motion  being  steady,  is  equal  to  the  angle  of  friction ;  call 
it  0. 


•} 

i, ; '    ( 
■  1     ■ 

Fig.  2S3- 


1    J 


I 

e  3. 


ii 


iii^ 


ri  I! 


330  THEORY  OF  STRUCTURES. 

Taking  moments  about  ^, 


Pp  —  Qp  —  Rr  s\n  (p^o (i) 


Also, 


R' -  P" -\- Q  ^  2PQ  cos  6 (2) 


Let/=  sin  0  = 


A* 


Eq.  (i)  may  now  be  written 


,  jj.  being  the  coefficient  of  friction. 


Pp-  Qp-  fRr  =  O.  .  . 
If  Pand  Q  are  parallel  in  direction, 

^  =  O  and  R  =  P-l-Q. 
Let  the  figure  represent  a  wheel  and  axle. 


(3) 


Let  P  be  the  effort  and  Q  the  weight  lifted,  the  directir  r. 
of  Pand  Q  being  parallel. 

Let  ^Fbe  the  weight  of  the  "  wheel  and  axle." 

Let  /?,  and  R^  be  the  vertical  reactions  at  the  bi.      iigs. 

Let/  be  the  radius  of  the  wheel. 

Let  q      "  "         "        axle. 

Let  r     "  "        "        bearings. 

Take  moments  about  the  axis.     Then 


Pp  —  Qq  —  R,r  sin  cp  —  R^r  sin  0  =  o. 


(4) 


TOOTHED   GEARING. 


331 


But 


Hence, 


R,-\-R,=  W+P+Q (5; 


Pp  -  Qg  =  (JV-{-PJr  Q)r  sin  0  =  (W+PJr  Q)/r, 


or 


P{p-/r)  =  Q{g  +  /r)+/lVr 

Efficiency. — In  turning  through  an  angle  6', 
motive  work  =  PpB, 
useful   work  =  Qq^, 

Qq^      Qq 


(6> 


.'.  efficiency       = 

Q 


Ppe  -  Pp ' 


and  the  ratio  j,-  is  given  by  eq.  (6). 

16.  Toothed  Gearing. — In  toothed  gearing  the  friction  is 
partly  rolling  and  partly  sliding,  but  the  former  will  be  disre- 
garded, as  it  is  small  as  compared  with  the  latter. 


Fig, 


ass- 


Let  the  pitch-circles  of  a  pair  of  teeth  in  contact  at  the 
point  B  '  ^uch  at  the  point  A  ;  and  consider  the  action  before 
reaching    the   line    of    centres    (9,6>, ,    i.e.,    along   the   arc  of 

approach. 


^it;?!-;. 


ili 


-y 


I 


11 


l> 


i      I 


^sf^sm  •-  ■ 


332 


THEORY  OF  STRUCTURES, 


The  line  /i/)'  is  nonnal  to  the  surfaces  in  contact  at  the 
point  /)' 

Let  R  be  the  resultai^t  reaction  at  //.  Its  direction,  the 
motion  being  steady,  makes  an  angle  0,  equal  to  the  angle  of 
friction,  with  /!/>. 

Let  ^  be  the  angle  between  (7,C7,  and  AB, 

Let  the  motive  force  and  force  of  resistance  be  respective!)- 
equivalent  to  a  force /*  tangential  to  the  pitch-circle  C?,  ,anil 
to  a  force  Q  tangential  to  the  pitch-circle  (>, . 

Let  ;•, ,  i\  be  the  radii  of  the  two  wheels. 

The  work  absorbed  by  friction  in  turning  through  the  small 
arc  d& 

-=^{P-Q)ds (1) 

Consider  the  wheel  (?, ,  and  take  moments  about  the  centre, 

Pi\  =:  K\r,  sin  (^  -  </>)  +  x  sin  0(,     .     .    .    (j) 

where  /IB  =  r. 

Similarly,  from  the  wheel  O^ 


Qr^  — R\r^  sin  {H  —  <f>)  —  X  sin  <p\.     ,     .    .    (3) 


Hence, 


Q      ?•:■)  (6-  </>)  -  -  sin  0 

'P~  X         "' 

sin  {0  —  0)  -f-  -  sin  0 


•         •         •         • 


(4) 


and  therefore 


P-Q=Q 


(-  "I — \v  sin  0 


{^ 


?\v\  (^  —  0) sin  0 

Hence,  the  work  absorbed  by  friction  in  the  arc  ds 

(^+;-)-»^sin0flJf 


=  <2 


.    .   .   (f'l 


sin  {0  -'  0)    -  '-  sin  0 


In  1 

leaving 


and  t'^c 


Th 


e  1 


arc  respc 

f^iven  by 

the  arc  o 

A[Tain 


i.e.,  if 

and  this  c 
Simpl( 

ci'liciency 
H  diffe 

with  r,  a 

urcd  from 
Hence 


TOOTHED   GEARING. 


333 


111  precisely  the  same  manner  it  can  be  shown  that,  after 
leaving  the  line  of  centres ,  i.e.,  in  the  arc  of  recess. 


Q      sin  (^  -J-  0) sin  0 

sin  {H-\-  0)-f-  -  sin  «/» 


and  t^c  work  absorbed  by  friction  in  the  arc  ds 


(7) 


Q 


.    (8) 


sin  {(i  -{-  </') sin  <f) 


Q 


The  ratio  -.,  and  the  loss  of  iv&rk  given  by  cqs.  (4)  and  (6) 

Q 
an;  respectively  greater  than  the  ratio  r",  ■aw(^\.\\c  loss  of  ivorh 

^;iven  by  eqs.  (7)  and  (iS),  and  therefore  it  is  advisable  to  make 
the  arc  of  approach  as  small  as  possible. 

Aijain,  by  eq.  (4),  motion  will  be  impossible  if 


i.e.,  if 


sin  (6^  —  0)  +      sin  0  =  o ; 
cot  0  =  cot  ^ 


r,  sin  6^' 


aiul  this  can  only  be  true  if  the  direction  of  A'  passes  through  6^, . 

Simple  approximpte  expressions  for  ..'c  /t;.y^  xvork  and 
jilicioncy  may  be  obtained  as  follows : 

/^  differs  very  little  from  90°,  and  x  is  small  as  compared 
with  r,  and  differs  little  from  the  corresponding  arc  s  meas- 
ured from  A. 

Hence  the  work  absorbed  by  friction  in  the  arc  ds 

=  (2  tan  0('-  ^--Xds  =  (2/<(;  +  ^\ds, 


i% 


m 


% 


(i.'i' 


!■ 


iii 


334  THEORY  OF  STRUCTURES. 

and  the  work  lost  in  arc  of  approach  j, 


The  useful  work  done  in  the  same  interval  =  Qs^ . 
The  counter-efficiency  (reciprocal  of  efficiency) 


(2i.  +  G^(^+^) 


r, 


r]  2 


=  i  +  /^C^+^jA  .  .  (10) 

Similarly  for  the  arc  of  recess  j, , 

the  lost  work  =<2yu(^+i)^,      .     .    (u) 

and  the  counter-efficiency  =  i  -j-  M[~  +  -  )~.      .    (12) 

•    1.                 27ry,       27rr, 
If  J,  =  J,  =  pitch  —  p  —  - —  = ,  «, ,  n^  being  the  nam- 

ber  of  teeth  in  the  driver  and  the  follower,  respectively,  the  ex- 
pressions for  the  lost  work  given  by  eqs.  (9)  and  (11)  are  iden- 
tical, and  those  foh  the  counter-efficiency  given  by  eqs.  (lo"* 
and  (12)  are  also  identical. 

Thus,  the  whole  work  lost  during  the  action  of  a  pair  of 
teeth 

=  ^>^(^+^)/' 03) 

'  1  '  a 

and  the  counter-efficiency 

'1      '  j'  - 
=:l+;i;r(~  +  i) (15) 

This  last  equation  shovvs  that  the  efficiency  increases  with 
the  number  of  teeth. 


\Mi ' 


EFFICIENCY  OF  MECHANISMS. 


335 


If  the  follower  is  an  annular  wheel,  —  —  — 


must  be  substi- 


Thus,  with  an  an- 


tuted  for  — \-—   in  the  above  equations. 

nular    wheel    the    counter-efficiency   is    diminished    and    the 
efficiency,  therefore,  increased. 

It  has  been  assumed  that  R  and  Q  are  constant,  as  their 
variation  from  a  constant  value  is  probably  small.  It  has  also 
been  assumed  that  only  one  pair  of  teeth  are  in  contact.  The 
theory,  however,  holds  good  when  more  than  one  pair  are  in 
contact,  an  effort  and  resistance,  corresponding  to  P  and  Q, 
being  supposed  to  act  for  each  pair. 

17.  Bevel-wheels. — Let  I  A,  IB  represent  the  develop- 
ments of  the  axes  of  the  pitch- 
circles  //, ,  //,  of  a  pair  of  bevel- 
wheels  when  the  pitch-cones  are 
spread  out  flat,  0, ,  O^  being  the 
corresponding  centres. 

The  preceding  formulae  will  ap- 
ply to  bevel-wheels,  the  radii  being 
OJ,  OJ,  and  the  pitch  being  meas- 
ured on  the  circumferences  I  A,  IB. 

18.  Efificiency  of  Mechanisms. 
— Generally  speaking,  the  ratio  of 
the  effort  P  to  the  resistance  Q  in  a. 
mechanism  may  be  expressed  as  r 
function  of  the  coefificient  of  fric- 
tion fx.     Thus, 

Q    =  Hm)' 

If,  now,  the  mechanism  is  moved  so  that  the  points  of 
application  of  P  and  Q  traverse  small  distances  Jx,  Ay  in  the 
directions  of  the  forces, 


\ 


(1 


the  efficiency  = 


PAx 


I    Ay 
Jd^Ax' 


3S^ 


THEORY  OF  STRUCTURES. 


Ay 
But  the  ratio  -j-^  depends  only  upon  the  geometrical  rcLi- 

tions  between  the  different  parts  of  the  mechanism,  and  will 
therefore  remain  the  same  if  it  is  assumed  that  i-i  is  zero.  In 
such  a  case  the  efficiency  would  be  perfect,  or  the  motive  vvoik 
{PAx)  would  be  equal  to  the  useful  work  {QAy),  and  therefore 


I.  In 


^  i 


I  = 


Hence,  the  efficiency 


I     Ay 
F{o)A^' 


2?1; 


TABLE  OF  COEFFICIENTS  OF  AXLE-FRICTION. 


i     '.  ii 


Bell-met.il  on  bell-raetal 

Brass  on  brass 

Br.iss  on  cast-iron 

Cast-iron  on  bell-metal 

C.ist-iron  on  brass  

Cast-iron  on  cast-iron 

Cast-iron  on  liRnum-vitae 

Lignum-vitae  on  cast-iron 

Litfnum-s'itoe  on  lignum-vitae... 

Wroutrht-iron  on  bell-mclal 

WroU(jht-iron  on  cast-iron 

Wrou({ht-iron  on  liffnum-vit.ie.   . 


c 

•a 

0 

a 

'fi    : 

.S.S 

>y 

£^ 

■a  3 

0-^ 

Q 

c 

.097 

.079 

.072 

•'94 

.161 

.075 

.194 

.075 

•»37 

.075 

.185 

.  I 
.116 

.251 

.189 

■075 
.075 

.187 

"5 

c 

w  o 

3  u 

C  3 
°  J 

u 


.040 
.054 

•  054 
.054 

.092 

■  '7 
.07 
,054 
•"54 


I 

U  3 


.065 


.09 


^3 


.14 
•  14 


EXAMPLES. 


337 


EXAMPLES. 

1.  In  a  pair  of  four-sheaved  blocks,  it  is  found  that  it  requires  a  force 
P  to  raise  a  \veifi;ht  5/",  and  a  force  5/''  to  raise  a  weigiit  15/^'.  Show 
that  the  j^enerid  relation  between  the  force  P  and  the  weight  W  to  be 
raised  is  given  by 

P  =  -IV-  P'. 

Find  the  efficiency  when  raising  the  weights  5/"  and  15/". 

2.  Find  tlie  mechanical  advantage  v/hen  nn  inch  bolt  is  screwed  up 
byaiS-in.  spanner,  the  elTective  diameter  of  the  nut  being  I'i  in.,  the 
diameter  at  the  base  of  the  thread  .84  in.,  £ iid  .  1 5  being  the  coclFicient  of 
friction. 

3.  A  belt,  embracing  one-half  the  circumference  of  a  pulley,  transmits 
10  H.  P.  ;  the  pulley  makes  30  revolutions  per  minute  and  is  7  ft.  iri 
diameter.     Neglecting  slip,  find  T\  and  T-i\  i-i  being  .125. 

4.  A  A-in.  rope  passes  over  a  6-in.  pulley,  tlie  diameter  of  the  axis  bointf 
i  in. ;  the  load  upon  the  axis  =  2  x  the  rope  tension.  Find  tlie  elhcieiicv 
ol  tiic  pulley,  the  coefficient  of  axle-friction  being  .08  and  the  coefficient 
for  slilTnoss  .47. 

Hence  also  deduce  the  efficiency  of  a  pair  of  three-sheaved  blocks. 

5.  If  the  pulleys  are  50  ft.  c.  to  c.  and  if  the  tight  is  three  times  the 
slack  tension,  find  the  length  of  the  belt,  the  coefficient  of  friction  being 
iand  the  diameter  of  one  of  the  pulleys  12  in. 

6.  Show  that  the  work  transmitted  by  a  belt  passing  over  a  pulley 


will  be  a  ma.ximum  when  it  travels  at  the  rate  of  4/  _  ft.  per  sec,  T% 

being  the  slack  tension  and  tn  the  mass  of  a  unit  of  length  of  the  belt. 

The  tight  tension  on  a  20-in.  belt,  embracing  one -half  the  circum- 
ference of  the  pulley,  is  1200  lbs.  Find  the  maximum  work  the  belt  will 
transmit,  the  thickness  of  the  belt  being  .2  in.  and  its  weight  .0325  lb. 
per  cubic  inch.     (Coefficient  of  friction  =  .28.^ 

7.  In  an  endless  belt  passing  over  two  pulleys,  the  lea.st  tension  is 
150  lbs.,  the  coeflicient  of  friction  .28,  and  the  angle  subtended  by  the 
arc  of  contact  148".  Find  the  greatest  tension.  The  diameter  of  the 
larger  wheel  is  7S  in.,  of  the  smaller  10  in.,  of  the  bearings  3  in.  Find 
the  efficiency.  A  lightening-pulley  is  mndc  to  press  on  the  slack  side 
of  the  belt.  Assuming  that  the  working  tension  is  to  the  coefficient  of 
elasticity  in  the  ratio  of  i  to  80,  find  the  increment  of  the  arc  of  contact 


I'fT'' 

'PI 

IP 

<i 

^ 

^^^1 

1 

!i 

[.  ^; 

J! 

fi 

s  ■  ■ 

■;|  )■ 

il'.v 

■  \         ■  ■    ■ 

■  ■       5  :■  ■ "  '1 
-  f  ,  :  r 

■  J 

;      !       - 

r; 

M 


^l 


t^^y** 

1 

j 

\'U 


M 


i    ill 


I-    '■■ 


338 


THEORY  OF  STRUCTURES. 


on  the  belt-piillcy,  the  tension  of  the  slack  side,  and  the  force  of  the 
tightening-pulley. 

8.  A  belt  weighing  i  lb.  per  lineal  foot,  connects  two42-in.  pulleys,  one 
making  240  revolutions  per  minute.  Find  the  limiting  tension  for  which 
work  will  be  transmitted.  Also  find  the  tight  and  slack  tensions  and 
the  efficiency  when  the  belt  transmits  5  horse-power.  Diameter  of  axle 
=  2  in.;  coefficient  of  friction  =  .28. 

9.  A  circular  saw  makes  1000  revolutions  per  minute  and  is  driven 
by  a  belt  3  in.  wide  and  \  in.  thick,  its  weight  per  cubic  inch  being  .0325 
lb.  The  belt  passes  over  a  lo-in.  pulley,  embracing  one-half  the  cir- 
cumference, and  transmits  6  H.  P.  Find  the  light  and  slack  tensions, 
the  coefficient  of  friction  being  .28. 

10.  A  flexible  band,  embracing  three-fourths  of  the  circumference  of 
a  brake-pulley  keyed  on  a  revolving  shaft,  has  one  extremity  attached  to 
the  end  A  of  the  lever  AOB,  and  the  other  to  \.\\g  fixed  point  O  (between 
A  and  B^  about  which  the  lever  oscillates.  The  pressure  between  the 
band  and  pulley  is  effected  by  a  force  applied  at  right  angles  to  the  lever 
at  the  end  B.  Show  that  the  time  in  which  the-axle  is  brought  to  rest 
is  about  2i  •^iines  as  great  when  revolving  in  one  direction  as  in  the 
opposite  (/  =  .2). 

11.  In  a  Prony-brake  test  of  a  Westinghouse  engine,  the  blocks  were 
fixed  to  a  24-in.  fly-wheel  with  a  6-in.  face,  and  the  balance-reading  was 
48  lbs.;  the  distance  from  centre  of  shaft  to  centre  of  balance,  measured 
horizontally,  was  30  in.,  and  the  number  of  revolutions  per  minute  was 
624.    Find  the  H.  P.  Ans.  14.3. 

12.  An  engine  makes  150  revolutions  per  minute.  If  the  diameter  of 
the  brake-pulley  is  45  in.  and  the  pull  on  the  brake  is  50  lbs.,  find  the 
B.  H.  P.  Ans.  2.67. 

13.  A  small  water-motor  is  tested  by  a  tail  dynamometer.  The  pul- 
ley is  18  in.  in  diameter;  the  weight  is  60  lbs.;  the  spring  registers  a 
pull  of  50  lbs.;  the  number  of  revolutions  per  minute  =  500.  Find  the 
B.  H,  P.  Ans.  f 

14.  The  power  of  sn  engine  making  «  revolutions  per  minute  is 
tested  by  a  Prony  brake  having  its  arm  of  length  r  connected  with  a 
spring-balance  which  registers  a  force  P.  The  arm  is  vertical  and  the 
weight  W  oi  the  brake  is  supported  by  a  stifT  spring  fixed  vertically 
below  the  centre  of  the  wheel.  What  error  in  B.  H.  P.  would  be  intnj- 
duced  by  placing  the  spring  x  ft.  away  from  the  central  position  } 

Ans.  —p — ,  B  being  the  B.  H.  P. 

15.  Find  work  absorbed  by  friction  per  revolution  by  a  pivot  3  in. 
long  and  carrying  6  tons,  its  upper  face  being  6  in.  in  diameter,  coeffi- 
cient of  friction  .04,  and  2  a  being  90°. 


EXAMPLES. 


339 


i6.  The  diameter  of  a  solid  cylindrical  cast-steel  pivot  is  2}  in.  Find 
the  diameter  of  an  equally  efficient  conical  pivot. 

17.  The  pressure  upon  a  4-in.  journal  making  50  revolutions  per  min- 
ute is  6  tons,  the  coefficient  of  friction  being  .05.  Find  the  n.umber  of 
units  of  heat  generated  per  second;  Joule's  mechanical  equivalent  of 
heat  being  778  ft. -lbs. 

18.  A  water-wheel  of  20  ft.  d;  meter  and  weighing  20,000  lbs.  makes 
10  revolutions  per  minute;  the  gudgeons  are  6  in.  in  diameter  and  the 
coefficient  of  friction  is  .1.  Find  the  loss  of  mechanical  effect  due  to 
friction.  If  the  motive  power  is  suddenly  cut  ofT,  how  many  revolutions 
will  the  wheel  make  before  coming  to  rest  ?  Ans,  ^^  H.  P. ;  10.9. 

19.  A  fly-wheel  weighing  8000  lbs.  and  having  a  radius  of  gyration  of 
10  ft.  is  disconnected  from  the  engine  at  the  moment  it  is  making  27 
levolutions  per  minute  ;  it  stops  after  making  17  revolutions.  Find  the 
coefficient  of  friction,  the  axle  being  12  in.  in  diameter.        Ans.  .2325. 

20.  A  railway  truck  weighing  12  tons  is  carried  on  wheels  3  ft.  in 
diameter  ;  the  journals  are  4  in.  in  diameter,  the  coefficient  of  friction 
^jj.  Find  the  resistance  of  the  truck  so  far  as  it  arises  from  the  friction 
of  the  journals.  .  Ans.  ^yi  lbs. 

21.  A  tramcar  wheel  is  30  in.  in  diameter,  the  axle  2 J  in.;  the  coeffi- 
cient of  axle-friction  .08,  of  rolling  friction  .09.  Find  the  resistance  pei 
ton.  Ans.  28.37  lbs. 

22.  A  bearing  16  in.  in  diameter  is  acted  upon  by  a  horizontal  force 
of  50  tons  and  a  vertical  force  of  10  tons ;  the  coefficient  of  friction  is  ■y'j. 
Find  the  H.  P.  absorbed  by  friction  per  revolution.       Ans.  .906  H.  P. 

23.  A  steel  pivot  3  in.  in  diameter  and  under  a  pressure  of  5  tons 
makes  60  revolutions  per  minute  in  a  cast-iron  step  well  lubricated  with 
oil.  How  much  work  is  absorbed  by  friction,  the  coefficient  of  friction 
being  .08  ? 

24.  A  pair  of  spur-wheels  are  4  in.  and  2  in.  in  diameter  ;  the  flanks 
of  the  teeth  are  radial ;  the  larger  wheel  has  16  teeth  ;  the  arc  of  ap- 
proach =  arc  of  recess  =  f  of  the  pitch.  Show  how  to  form  the  teeth, 
and  find  their  efficiency.    (Coefficient  of  friction  =  .11.) 

25.  Find  the  work  lost  by  the  friction  of  a  pair  of  teeth,  the  number 
of  teeth  in  the  wheels  being  32  and  16,  and  the  diameter  of  the  larger 
wheel,  which  transmits  3  horse-power  at  50  revolutions  per  minute,  3  ft. 

26.  The  driver  of  a  pair  of  wheels  has  120  teeth,  and  each  wheel  has 
an  addendum  equal  to  .28  times  the  pitch  ;  the  arcs  of  approach  and 
recess  are  each  equal  to  the  pitch  ;  the  tooth-flanks  are  radial.  (Coeffi- 
cient of  friction  =  .  106.)     Find  the  efficiency.     ' 


"^Pl 

HI 

! 

(■!iU 

, 

.'i' 

1 

I      |.[^.:.* 


P  -^f^'i'l 


jl 

- 1  ( 


k:lkl>ii.>mj.  ^  uf 


i 


Ill 


'ill. 


p' 


Fig.  2SJ, 


Fig.  258. 


CHAPTER   VI. 

ON  THE  TRANSVERSE  STRENGTH  OF  BEAMS. 

I.  To  determine  the  Elastic  Moment. — Let  the  plane  of 

R  the  paper  be  a  plane  of  symmetry 
Avith  respect  to  the  beam  PQRS. 
If  the  beam  is  subjected  to  the 
action  of  external  forces  in  this 
plane,  PQRS  is  bent  and  as- 
sumes a  curved  form  P'Q'R'S'. 
The  upper  layer  of  fibres,  Q'R',  is 
extended,  the  lower  layer,  P'S',  is  compressed,  while  of  the 
layers  within  the  beam,  those  nearer  P'S'  are  compressed  and 
those  nearer  Q'R'  are  extended.  Hence,  there  must  be  a  layer 
M'N'  between  P'S '  and  Q'R'  which  is  neither  compressed  nor 
extended.  It  is  called  the  neutral  surface  (or  cylinder),  and 
its  axis  is  perpendicular  to  the  plane  of  flexure.  In  the  present 
treatise  it  is  proposed  to  deal  with  flexure  in  one  plane  only, 
and,  in  general,  it  will  be  found  more  convenient  to  refer  to 
M'N'  as  the  neutral  line  (or  axis),  a  term  only  used  in  refer- 
ence to  a  transverse  section. 

If  a  force  act  upon  the  beam  in  the  direction  of  its  length, 
the  lower  layer  P'S' ,  instead  of  being  compressed,  may  be 
stretched.  In  such  a  case  there  is  no  neutral  surface  within 
tthe  beam,  but  theoretically  it  still  exists  some- 
where ivithout  the  beam. 

Let  ABCD  be  an  indefinitely  small  rect- 
angular element  of  the  unstrained  beam,  and 
let  its  length  be  s.  Let  A'B'C'D',  Fig.  260, 
be  the  element  after  deformation  by  the  external  forces. 

340 


/ 

B 

'1 

p 

p 
D 

Q 
C 

Fig.  359. 


%\ 


Ilia 


THE  ELASTIC  MOMENT. 


P'Q',  the  neutral  line,  being  neither  com- 


341 


A'. 


pressed  nor  extended,  is  unchanged  in  length  p! 
and  equal  to  PQ  —  s. 

Let  the  normals  at  P'  and  Q'  to  the  neutral 
line  meet  in  the  point  0  \  O  h  the  centre  of 
curvature  of  P'Q'. 

Also,  as  the  flexure  of  the  element  is  very 
small,  the  normal  planes  through  OP'  and  OQ' 
may  be  assumed  to  be  perpendicular  to  all  the 
layers  which  traverse  the  corresponding  sec- 
tions of  the  beam,  so  that  they  must  coincide 
with  the  planes  A'D'  and  B'C,  respectively. 

The  assumptions  made  in  the  above  are  : 

(^2:)  That  the  beam  is  symmetrical  with 
respect  to  a  certain  plane. 

[b)  That  the  material  of  the  beam  is  homo- 
geneous. 

(r)  That  sections  which  are  plane  before  bending  remain 
plane  after  bending. 

{d)  That  the  ratio  of  longitudinal  stress  to  the  correspond- 
ing strain  is  the  ordinary  (i.e.,  Young's)  modulus  of  elasticity 
notwithstanding  the  lateral  connection  of  the  elementary 
layers. 

{c)  That  these  elementary  layers  expand  and  contract 
freely  under  tensile  and  compressive  forces. 

Consider  an  elementary  layer  p'q',  of  length  s',  sectional 
area  a^ ,  and  distant  j,  from  the  neutral  surface. 

-LctOP'  =R=  OQ'. 

From  the  similar  figures  OP'Q'  and  Op'q' , 


Op'        p'q' 


OP' ~  P'Q' 


or 


R±_y. 

R 


=  — ,     and  therefore     ^  = 


R 


s'  —  s 


Also,  if  /,  is  the  stress  along  the  layer /»'^', 


j    ■ 

\il 

^     S'  —S  r-       f^  ^ 


342 


THEORY  OF  STRUCTURES. 


n  > 


E  being  the  coefificient  of  elasticity  of  the  material    of_the 
beam. 

So,  il  f,,  a^,  y^,  it,  a^,  jt,  .  .  .  are  respectively  the  stress, 
sectional  area,  and  distance  from  the  neutral  surface,  of  the 
several  layers  of  the  element, 


_E  _E 


The  total  stress  along  the  beam  is  the  algebraic  sum  of  a 
these  elementary  stresses, 


E  E 


R> 


R 


Again,  the  moment  of  /,  about  P'  =  t^y^  —  —  a^y^  ; 

i 

E 

it  U  ««        /  "  «       /     ^ ^      ..  3  . 

*«  —  *a  /a  —    r>  "a  J'a    > 


R 


E 


u  u  n    4        ««        "    =  /  1/ n   V '  • 


and  so  on. 

Thus,  the  Elastic  Moment  for  the  section  A' D'  —  the  alc^e. 
braic  sum  of  the  moments  of  all  the  elementary  stresses  in  the 
different  layers  about  P' , 


=  ^ijV.  +  t^y^  +  t,y^  4-  .  .  .  =  T,-(«, y^  +  «,/»'  +  ••  •) 


R 

E 
R 


=  %^iccf)' 


I'l 


THE  ELASTIC  MOMENT. 


343 


Now,  2  («y)  is  tbe  tnoment  of  inertia  of  the  section  of  the 
beam  through  A'D\  with  respect  to  a  straight  Hne  passing 
through  the  neutral  Hne  and  perpendicular  to  the  plane  of 
flexure,  i.e.,  the  plane  of  the  paper.  It  is  usually  denoted  by 
/  or  Ak'\  A  being  the  sectional  area,  and  k  the  radius  of 
gyration.     Thus, 

E         E 
the  elastic  moment  =  -  /=  -^Ak^, 

But  the  elastic  moment  is  equal  and  opposite  to  the  bending 
moment  {M)  due  to  the  external  forces,  at  the  same  section. 
Hence 

~I=.~Ak^  =  M. 


the  alc;e- 
es  in  the 


Note. — It  is  necessary  in  the  above  to  use  the  term  alge- 
braic, as  the  elementary  stresses  change  in  character,  and 
therefore  in  sign,  on  passing  from  one  side  of  the  neutral  sur- 
face to  the  other. 

Cor.  I.  Bearing  in  mind  assumption 
(<?),  the  figure  represents  on  an  exaggerated 
scale  the  transverse  section  of  the  beam 
at  A'D',  the  upper  and  lower  breadths 
of  the  beam,  ^'^"  and  D'D",  being  re- 
spectively contracted  and  stretched,  and 
being  also  arcs  of  circles  having  a  common 
centre  at  O'. 

Let  R'  be  the  radius  of  the  arc  P'P", 
whose  length  remains  unchanged. 

Let   mE  be  the  lateral  coefficient  of 
elasticity,  in  being  a  numerical  coefficient. 
As  before,  for  any  layer  at  a  distance  j/  from  P'P", 


Fig.  261. 


mE       t 
R'  ~  ay~ 

E 
R' 

'.  R'  =  mR. 

-iM 


U' 


THEORY  OF  STRUCTURES. 


Thus,  ivithin  the  limits  of  elasticity,  the  curvature  of  the  breadth 

is  —  that  of  the  length,  and  does  not  sensibly  affect  the  re- 
in 

sistance  of  the  beam   to  bending.     The   influence,  however, 

upon  the  bending  may  become  sensible  if  the  breadth  is  very 

large  as  compared  with  the  depth,  as,  e.g.,  in  the  case  of  iron 

or  steel  plates. 

Cor.  2.  If  the  resolved  part  of  the  external  forces  in  the 

direction  of  the  length  of  the  beam  is  tiil, 

E 
the  total   longitudinal  stress  =  -zr2{a}>)  =  o,  or  2{a}')  =  o, 


showing  that  P'  must  be  the  centre  of  gravity  of  the  section 
through  A'D'.  Hence,  when  the  external  forces  produce  no 
longitudinal  stress  in  the  beam,  the  neutral  line  is  the  locus  of 
the  centres  of  gravity  of  all  the  sections  perpendicular  to  the 
length  of  the  beam. 

Cor.  3.  If  /,  a,  y  be,  respectively,  the  stress,  sectional  area, 
and  distance  of  a  fibre  from  the  neutral  line,  then 


-^ay  =  t,   or    -^ 


jj/  =  -  =  intensity  of  stress  =-fy,  suppose, 


fyE.E 


/=M==^'-/. 


Example  i.  A  timber  beam,  6  in.  square  and  20  ft.  long, 
rests  upon  two  supports,  and  is  uniformly  loaded  with  a  weight 
of  1000  lbs.  per  lineal  foot.  'Determine  the  stress  at  the  centre 
at  a  point  distant  2  in.  from  the  neutral  line. 

Also  find  the  central  curvature,  E  being  1,200,000  lbs. 


I—-^—  —  108,    M  =^  looa  X  10—  1000  X  5=  5000   ft.-lbs. 
12 

=  60,000  inch-lbs.,     and    y  ■=  2  in. 


and  the  stress  in  the  metal  is  necessarily  greatest  at  the  central 
section. 


W-\-  5200 

M,  at  the  centre,  =  — .  30.  12  inch-lbs. ; 

8 


/  ''2  1 

/,  =  2  X  2240  lbs.,     and     -  =  ?rrV  =  —  .  15'..  -. 


^HasMiii 


'Jtic  !i  i|'".'f 


INTERNAL   STRESSES.  345 

Hence  from  the  above  equations, 


I20OO00  „        ^  /y      „ 

— 73—  X  108  =  60000  =  —108. 

K  2 


Thus  R  =  2160  in.  =  180  ft.,     and    ^  =  11  ii-^  lbs.  per  sq.  in. 

Ex.  2.  A  standpipe  section,  33  ft.  in  length  and  weighing 
5720  lbs.,  is  placed  upon  two  supports  in  the  same  horizontal 
plane,  30  ft.  apart.  The  internal  diameter  of  the  pipe  is  30 
in.,  and  its  thickness  .V  inch.  Determine  the  additional 
uniformly  distributed  load  which  the  pipe  can  carry  between 
the  bearings,  so  that  the  stress  in  the  metal  may  nowhere  ex- 
ceed 2  tons  per  square  inch. 

Let  W  be  the  required  load  in  pounds. 

30 
The  weight  of  the  pipe  between  the  bearings  =  —  .  5720 

—  5200  lbs. 

Thus,  the  total   distributed  weight  between  the  bearings 

=  (rr+ 5200  lbs.) 


Now 


c 


W-\-  5200  22      ,   I 

~ .  30.  12  =  2  .  2240  .— .  15'.  -  =  72000  X  22, 

o  72 


and  hence  W  ^=  30,000  lbs. 


■}■.  i- 


ii 


346 


THEORY  OF  STRUCTURES. 


Cor.  4.  The  beam  is  strained  ♦"o  the  limit  of  safety  when 
either  of  the  extreme  layers  A' B\  D'C  is  strained  to  the  limit 

of  elasticity.     In  such  a  case,  the  least  of  the  values  of  —  for 

y 

the  extreme  \s.ycrs  A' B' ,  D' C  is  the  greatest  consistent  with 
the  strength  of  the  beam  ;  and  \i  f^  and  c  are  the  corresponding 
intensity  of  stress,  and  distance  from  the  neutral  axis, 


R  c 


EXAMTLE. — Compare  the  strengths  of  two  similarly  loaded 
beams  of  the  same  mai°rial,  of  equal  lengths  and  equal  sectional 
areas,  the  one  being  round  and  the  other  square. 

Let  r  be  the  radius  of  the  round  beam  ;  /,.,  the  intensity  of 
the  okin  stress. 

Let  a  be  a  side  of  the  square  beam;  _/,,  the  intensity  of 
the  skin  st.'ess.     Then 


TTr'  =  rt' ;    /,  for  round  bar,  = ,  and  for  square  bar  =  --. 

4  '  12 


Also,  since  the  beams   are  similarly  loaded,  the   bending 
moments  at  corresponding  points  are  equal. 


prcs 

l^ut 

stros 

shea 

diffc 

with 


so  that 


"  r    ^ 

M 

a  12' 

2 

fr       2    a'        2 
/.       3  ^^"       3 

V 

/22               /88 

7  ~  V  63 

Thus,  under  the  same  load,  the  round  beam  is  strained  to 
a  greater  extent  than  the  square  beam,  and  the  latter  is  the 
stronger  in  the  ratio  of  i  88  to   V63. 


B/i  EA  KIA'G    WEIGH  TS. 


347 


Cor.  5.  The  neutral  surface  is  neither  stretched  nor  com- 
pressed, so  that  it  is  not  subjected  to  any  longitudinal  stress, 
liut  it  by  no  means  follows  that  this  surface  is  wholly  free  from 
stress,  and  it  will  be  subsequently  seen  that  the  effect  of  a 
shearing  force,  when  it  exists,  is  to  stretch  and  compress  the 
different  particles  in  diagonal  directions  making  angles  of  45° 
with  the  surface. 


Cor.  6.  For  a  rectangular  beam,  /  = 


12 


and     c  ■=■ 


c 


-7 =  T-bir 

a    12       6 


If  the  beam  is  fixed  at  one  end  and  loaded  at  the  other 
with  a  weight  IV,  the  maximum  bending  moment  =  IV/. 

If  the  beam  is  fixed  at  one  ei:d  and  loaded  uniformly  with 
a  weight  wl  =  W,  the  maximum  bending  moment 


2 


m 

2 


If  the  beam  rests  upon  two  supports  and  carries  a  weight  W 

IVl 
at  the  centre,  the  maximum  bending  moment  =::  — — . 

If  the  beam  rests  upon  two  supports  and  carries  a  uniformly 
distributed  load  of  ivl  =  W,  the  maximum  bending  moment 

-  '8   ~    8"* 

f  1.  jt 
Hence,  in  the  first  case,  W  =  y  —j~\ 


"    "    second  "     W 


6^  r 


"     "    third     "    JF=^4y; 

"    '•    fourth  "     ]V=  i^'T- 

6      I 


(!'     I 


I 


yi 


348 


In  general, 


THEORY  OF  STRUCTURES. 


^" 


il. 


q  being  some  coefificient  depending  upon  the  manner  of  the 
loading. 

Now,  if  the  laws  of  elasticity  held  true  up  to  the  point  of 
rupture,  these  equations  would  give  the  breaking  weights  (JF), 
corresponding  to  different  ultimate  unit  stresses  (/),  but  tlic 
values  thus  derived  differ  widely  from  the  results  of  experi- 
mert.     It   is  usual  to  determine  the  breaking  weight  {W)  of  a 

rectangul       beam  from  the  formula  W  ^^  ^  ~7~'  where  C  is  a 

constant  which  depends  both  upon  the  manner  of  the  loading 
and  the  nature  of  tlie  material,  and  is  called  the  coefficient  of 
rupture. 

The  modulus  of  rupture  is  the  value  of/"  in  the  ordinary 

bcnding-moment  formula  (J/  —  -/)   when   the   load  on  the 

beam  is  its  breaking  load. 

The  preceding  equations,  however,  may  be  evidently  em- 
ployed to  determine  the  breaking  weights  in  the  several  cases 

by  making  "^(7  =  C.     In  this  case /is  no  longer  the  real  stress, 

but  may  be  called  the  coefficient  of  bending  strength. 

The  values  of  C  for  iron,  steel,  and  timber  beams,  supported 
at  the  two  ends  and  loaded  in  the  centre,  are  given  in  the 
Tables  at  the  end  of  Chapter  III. 

The  corresponding  value  of  /  is  obtained  from  the  equation 


or 


f-.--\C. 


Example. — Determine   the  central  breaking  weight  of  a 


* 

\\i 

EQUALIZATION  OF  STRESS. 


349 


red-pine  beam,  lo  in.  deep,  6  in.  wide,  and  resting  upon  two 
supports  20  ft.  apart. 

The  value  of  C  for  red  pine  is  about  5700.     Hence, 


the   breaking  weight  =  IV  =  5700 


6.10' 


20  X  12 


-  =  14,250  lbs. 


2.  Equalization  of  Stress. — The  stress  at  any  point  of  a 
beam  under  a  transverse  load  is  proportional  to  its  distance 
from  the  neutral  plane  so  long  as  the  elastic  limit  is  not  ex- 
ceeded. At  this  limit  materials  which  have  no  ductility  give 
way.  In  materials  possessing  ductility,  the  stress  may  go  on 
iiicieasing  for  some  distance  beyond  the  elastic  limit  witliout 
prixl.'cing  rupture,  but  the  stress  is  no  longer  proportional  to 
C  iistance  from  the  neutral  plane,  its  variation  being  much 
slower.  This  is  due  to  the  fact  that  the  portion  in  compres- 
sion acquires  increased  rigidity  and  so  exerts  a  continually 
increasing  resistance  (Chap.  Ill)  almost  if  not  quite  up  to 
the  point  of  rupture,  while  in  the  stretched  portion  a  flow  of 
metal  occurs  and  an  appro.ximately  constant  resistance  to  the 
stress  is  developed.  Thus,  there  will  be  a  more  or  less  perfect 
equalization  of  stress  throughout  the  section,  accompanied  by 
an  increase  of  the  elastic  limit  and  of  the  apparent  strength, 
the  increase  depending  both  upon  the  form  of  section  and  the 
ductility. 

For  example,  if  the  tensile  elastic  limit  is  the  same  as  the 
compressive,  the  shaded  portion  of  Fig.  262  gives  a  graphical 


"W 


Fig.  26a. 


Ban 

-  ' I 

Fig.  263. 


Fig.  264. 


representation  of  the  total  stress  in  a  beam  of  rectangular 
section  when  the  straining  is  within  the  elastic  limit.  Beyond 
this  limit,  it  may  be  represented  as  in    Fig.  263,  and  will   be 


it' 


J' ) 


n 


If  I' 


350 


THEORY  OF  STRUCTURES. 


intermediate  between  Fig.  262  and  the  shaded  rectangle  of 
Fig.  264  which  corresponds  to  a  state  of  perfect  equaliza- 
tion. 

3.  Surface  Loading.* — It  may  be  well  to  draw  attention  to 
another  important  assumption  upon  which  is  based  the  matlie- 
matical  treatment  of  the  problem  of  Beam  Flexure. 

It  has  been  assumed  that  the  external  forces  acting  on  a 
beam  can  be  so  applied  that  they  may  be  considered  as  dis- 
tributed uniformly  over  the  whole  section.  Thus  when  a  beam 
encastr^  is  loaded  at  the  free  end,  Fig.  265,  the  load  P  is  as- 


!b 


Fig.  2*5. 

sumed  to  be  uniformly  distributed  over  the  section  ab^  i.e., 
each  element  in  the  section  is  supposed  to  experience  the  same 
amount  of  strain  due  to  the  load,  and  the  reaction  of  the  wall 
is  also  supposed  to  be  uniformly  distributed  over  each  element 
in  the  section  cd. 

It  is  clear  that  such  suppositions  must  be  far  from  the 
truth. 

In  practice,  the  load  P  must  be  hung  by  some  means  from 
the  beam,  say  by  a  stirrup  passing  over  the  top.  The  whole 
load  is  then  concentrated  at  the  line  of  contact  of  the  stirrup 
with  the  beam,  and  it   is   obviously  untrue  to  say  that  every 

*  This  aiiicle  was  kindly  written  by  Professor  Carus-Wilson  and  is  an 
abstract  ol  a  Paper  presented  by  hiin  to  the  Ptiysiiai  Society, 


SURFACE  LOADIXG. 


351 


element  in  the  section  ab  is  equally  strained.  But  more 
than  this.  It  has  been  assumed  that,  taking  the  effect  of 
the  load  as  distributed  uniformly  over  the  section  ab,  and 
a  certain  deflection  thereby  produced,  the  effect  of  P  on 
each  element  of  the  section  ab  may  be  disregarded  in  com- 
parison with  the  strains  involved  in  the  deflection  whicla  P 
produces. 

It  will  probably  be  difificult  at  first  to  grasp  the  fact  that 
certain  measurable  effects  have  been  actually  neglected,  but 
that  this  is  so  may  be  seen  by  supposing  the  beam  in  question 
to  be  a  pine  beam,  and  the  stirrup  of  iron.  Experience  proves 
that  with  a  very  moderate  load  the  beam  will  be  indented  at  a. 
But  the  theory  shows  that  the  longitudinal  tension  at  a  is 
zero  and  increases  to  a  maximum  at  d. 

Thus,  so  far  from  the  squeezing  effect  of  the  load  being 
distributed  uniformly  over  the  section  ab,  it  is  concentrated  at 
rt,  and  hence  it  is  impossible  to  neglect  it- 
Engineers  have  always  recognized  the  existence  of  this 
"surface-loading"  effect  in  practice,  and  where  possible,  have 
provided  a  good  "  bearing"  in  order  to  a^^oid  such  local 
strains :  but  this  cannot  always  be  done — as,  for  instance, 
in  the  case  of  rollers  under  bridge  ends.  The  theory  of  flex- 
ure is  therefore  manifestly  incomplete  if  it  cannot  take  into 
account  the  actual  manner  in  which  the  loads  are  and  must  be 
applied. 


■( 


t 


Fig.  266. 


It  can  be  shown  that  the  effect  of  placing  a  pressure  of  p 
tons  per  inch  run,  say  in  the  form  of  a  loaded  roller,  on  a  beam 
resting  upon  a  flat  surface,  as  in  Fig.  266,  to  prevent  it  from 


! 


I  I 


y ! 


i\ 


352 


THEORY  OF  STRUCTURES. 


bending,  is  to  compress  every  element  say  along  ab  with  an 
intensity  given  approximately  by  the  equation 


r_2p[\       _X\ 


where /is  the  pressure  at  a  distance  x  from  a^  the  point  of  con- 
tact, and  h  =  ab.  This  is  the  equation  to  a  curve  be  which  is 
approximately  an  hyperbola. 

When  a  beam  is  bent  by  the  application  of  external  forces,  a 
very  close  approximation  to  the  true  condition  may  be  obtained 
by  superposing  this  surface-loading  effect  on  that  found  for 
bending. 

Take  the  case  of  a  beam  supported  at  the  ends  and  loaded 
at  the  centre,  and  let  it  be  required  to  find  the  condition  along 
ab,  Fig.  267. 


Fig.  267. 

The  effect  of  the  bending  is  to  produce  compression  above 
and  tension  below  the  point  f,  and  these  effects  may  be  repre- 
sented by  a  right  line  dc  passing  through  c. 

The  surface-loading  effect  may  be  represented  by  an  hyper- 
bola giving  the  compression  at  any  point  along  ab  due  to  tlic 
load.     The  hyperbola  and  straight  line  will  intersect  in   two 


S  UKFA  CE   1. 0.1  DING. 


3ii 


poin  :s  h  and/,  which  shows  that  at  two  points  h' ,f'  along ^'(5  the 
vertical  squeeze  produced  by  the  load  is  of  equal  intensity  to 
the  horizontal  squeeze  produced  by  the  bending;  hence  aii 
cl  -inent  at  each  of  these  points  is  subject  to  cubical  compres- 
Mon  only.  From  a  to/'  the  beam  is  squeezed  vertically,  from 
/  to  li'  it  is  squeezed  horizontally,  and  from/;'  to  b  it  is  stretched 
horizontally.  The  intensities  are  given  at  every  point  by  the 
difference  between  the  ordinates  of  the  line  of  bending  dc  and 
the  curve  of  loading.  It  will  appear  that  one  effect  of  surface- 
loading  is  to  make  the  neutral  axis  rise  up  under  the  load  and 
pass  through  the  point  h\  for  there  is  neither  compression  nor 
tension  at  that  point. 

This  can  be  verified  by  examining  the  condition  of  a  bent 
glass  beam  by  polarized  light.  The  neutral  axis  is  pushed  up 
under  the  load  and  there  is  a  black  ring  passing  through  the  point 
/'.  If  the  span  i  diminished  and  the  load  kept  constant,  it  is 
clear  that  ae  will  become  less,  while  the  curve  of  loading  remains 
the  same,  until  the  line  dec  ceases  to  cut  the  curve  ;  every 
clement  along  ah  will  then  be  subjected  to  horizontal  stretch, 
and  the  stretch  is  greatest  at  a  ;  the  result  obtained  by  neglect- 
ing the  surface  loading  is  that  only  elements  from  c  to  b  are 
stretched,  the  greatest  stretch  being  at  b.  The  position  of  the 
"  neutral  points  "  is  given  by  the  equation 


h      4  i6      m 


where  y  is  the  distance  from  the  top  edge,  h  equals  the  depth 
ah.  111  =  —. 4,  and  a  =  one-half  of  the  span. 

For  all  elements  in  ab  to  be  stretched,  the  ratio  of  span  to 

depth,  viz.,  ^^,  must  be  equal  to  or  less  than  4.25.     In  other 

words,  for  any  beam,  and  any  load,  if  the  span  is  less  than  4^ 
times  the  depth,  every  element  in  the  normal  under  the  load 
is  ^.tretched  horizontally. 


*••*, 

354 


THEORY  OF  STRUCTURES. 


4.  Beam  acted  upon  by  a  Bending  Moment  in  a  Plane 
which  is  not  a  Principal  Plane. 

Let  XOX,  YO  V  be  the  principal  axes  of  the  plane  section 
of  the  beam. 


Fig.  368. 


Let  the  axis  MOM  of   the  bending  moment  M  make  an 
angle  a  with  OX. 

M  may  be  resolved  into  two  components,  viz., 


M  cos  a  z=i  X    and     M  sin  «  =r  F. 

These  components  may  be  dealt  with  separately  and  the 
results  superposed. 

Thus,  the  total  stress,/,  at  any  point  (x,y) 

=  stress  due  to  .^  -|-  stress  due  to  F  =  -  -  -f-  -7-  =  /, 

I., ,  ly  being  the  moments  of  inertia  with  respect  to  the  axes 
XOX,  YO  Y,  respectively. 

If  the  point  {xy)  is  on  the  neutral  axis,  then 

Xy   .    Yx 

■'■X  -"jl 


or 


tan/8  = 


-y  _ 


fi  being  the  angle  between  the  neutral  axis  and  XOX. 


ifli^^ 


SPRINGS. 


355 


Also  see  Art.  6,  Chap.  VIII.  In  this  article  ^  is  the  angle 
between  the  neutral  axis  and  the  axis  of  the  couple,  i.e.. 
d-  (i  —  a. 

5.  Springs. — {a)  Flat  Springs. — If  two  forces,  each  equal 
to  P  but  acting  in  opposite  directions  in  the  same  straight  line, 
are  applied  to  the  ends  of  a  straight  uniform  strip  of  flat  steel 
spring,  the  spring  will  assume  one  of  the  forms  shown  below, 
known  as  the  clastic  ctirvc.  This  curve  is  also  the  form  of  the 
linear  arch  best  suited  to  withstand  a  fluid  pressure,  Chap. 
XIII. 


Consider  a  point  B  of  the  spring  distant  y  from  the  line  of 
action  of  P.     Then 

FlI 
Py  —  bending  moment  at  ^  =  -zy- , 


r"Tr 


356 


THEORY  OF  STRUCTURES. 


Ill' , 


i)  S  '.■: : 


R  being  the  radius  of  curvature  at  B,  and  /  the  moment  of 
inertia  of  the  section. 

If  E  and  /are  both  constant, 


Rf 


a  constant 


is  the  equation  to  the  elastic  curve. 

{d)  Spiral  Springs  (as,  e.g.,  in  a  watch). — Let  the  figure  rep- 
resent a  spiral  spring  fixed  at  C  and  to  an  arbor  at  A,  ami 

subjected  at  every  point  of  its  length 
to  a  bending. action  only. 
Q  ]     I  Consider    the    equilibrium    of    any 

4^^J  portion  AB  of  the  spring. 

A'  ^^-Q  The  forces  at  A  are  equivalent  to  a 

B  ^p      couple  of  moment  M,  and  to  a  force  P 

^'°-  '7^-  acting  in  some  direction  AD. 

This  couple  and  force   must   balance  the  elastic  moment 
at.^. 

.-.  M-{-  Py  =  £1  X  change  of  curvature  at  B, 
y  being  the  distance  of  B  from  the  line  of  action  of  P,  or 


M- 


R^  being  the  radius  of  curvature  at  B  before  winding,  and  R 
that  after  winding. 

Let  ds  be  an  elementary  length  of  the  spring  at  B. 

Then,  for  the  whole  spring, 

2(/./+  Py)ds  =  £/:s(f  -  ~)  =  E!2{dd  -  de;), 

6r    M2ds  -{-  P2yds  —E/X  total  change  of  curvature  between 
A  and  C ;  , 

.'.Ms-YPsy^EI{B-  6/„), 


SPJilJVGS. 


357 


s  being  the  length  of  the  spring,  _y  the  distance  of  its  C.  of  G. 
from  AD,  B  the  angle  through  vvhicii  the  spring  is  wound  up, 
and  B^  the  "  unwinding"  due  to  the  fixture  at  C.  With  a  large 
number  of  coils  the  distance  between  the  C.  of  G.  and  A  may 

i)e  assumed  to  be  nil  and  then  y  =  o. 

Also,  if  the  spring  is  so  secured  that  there  is  no  change  of 
direction  relatively  to  the  barrel, 

6',  —  o,     and     Ms  =  Eld. 

Let  the  winding-up  be  effected  by  a  couple  of  moment 
Qq  =  M,  Q  being  a  tangential  force  at  the  circumference  of  a 
circle  of  radius  q. 

The  distance  through  which  Q  moves  (or  deflection  of  Q) 

=  ge  =  %s,    since    M  =  -L 
^        ch  c 


/being  the  skin  stress,  and  c  the  distance  of  the  neutral  axis 
of  the  spring  from  the  skin. 

Thus,  if  b  is  the  width  of  a  spring  of  circular  or  rectangular 

section,  c  =■  — .  and  hence 
'  2 

"zqf 
the  deflection  =  -rr-S' 
bE 

The  work  done  =  -Q  x  deflection  = g6  = 

2  2  q^  2 


2    Ec'    ~    2       Ec'     ~    2E      C 


5"  > 


/'  being  the  square  of  the  radius  of  gyration,  A  the  sectional 

area  of  the  spring,  and  V  its  volume. 

k'        I 
In  case  of  spring  of  rectangular  section  -p  =  -  . 


"    circular 


^ 
c' 


I 

4 


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358 


THEORY  OF  STRUCTURES. 


Again,  the  spiral  spring  in  Fig.  277  is  wholly  subjected  tf> 
a  bending  action  by  means  of  a  twisting  couple  of  moment 
M  z=i  Qq'xn  a  plane  perpendicular  to  the  axis  of  the  spring. 
Any  torsion  in  the  spring  itself  is  now  due  to  the  coils  not 
being  perfectly  flat. 


((       t< 


"•  Fig.  J77. 

Let  /?,  =  radius  of  a  coil  before  the  couple  is  applied. 

^  being  che  angle  of  twist ;  or      .   •        ,        . 

Qqs       Ms        s         s 
...      .:    .  -^  =  ^  =  ^-  -  Z  =  (^-  ^•)2'^' 


JV being  the  number  of  coils  before  the  couple  is  applied,  and 
X       <<  <<         <<      <<     after       "        "        "       " 

The  distance  through  which  Q  acts,  i.e.,  the  "  deflection,'* 


N,    "       " 


=^''=^ 


and  the  work  done 


"2        2E  c"' 


I  rv 

—  >-  —p~  for  spring  of  rectangular  section, 

I  rv 

-  2,   E 


"    circular 


6.  Beams  of  Uniform  Strength. — A  beam  having  the 
same  maximum  unit  stress  (/)  at  every  section  is  said  to  be  a 
beam  of  uniform  strength. 


s 


BEAMS  OF   UNIFORM  STRENGTH. 


359 


At  any  section  of  a  beam  An{=:l)  denote  the  bending 
moment  by  M,  the  depth  of  the  beam  by  y,  and  its  breadth 
by  b.    Then 

ll=  M=  ^Ak\ 
c  c 


c  being  the  distance  of  the  skin  from  the  neutral  axis,  and  A 
the  area  of  the  section. 

Evidently  c  and  k  are  each  proportional  toj/,  and  A  to  by. 


or 


nfbf  =  M, 


J".il' 


«  being  a  coefficient  whose  value  depends  upon  the  form  of 
section. 

Four  cases  will  be  considered. 

Case  a.  Assume  that  the  breadth  b  is  constant,  and  let 


nfb  =  -.    Then 


or 


y  —  ±  s/JM. 


Thus  AB  may  be  either  the  lower  edge  of  the  beam,  the 
ordinates  of  the  upper  edge  being  the  different  values  of  y,  or 
it  may  be  a  line  of  symmetry  with  respect  to  tlic  profile,  in 

y 
which  case  the  ordinates  are  the  different  values  of  ±   -. 

Example  i.  A  cantilever  AB  loaded  at  the  free  end  zvith  a 
wcig/it  \V,. 

At  a  distance  x  from  A, 

f  =  pM  =  p  W\.x, 


Theoretically,    therefore,    the 
beam,  in    elevation,    is   the    area    "^ 
ACD.   the    curve    CAD   being    a 


Pic.  178. 


'.'     I 


<.  ! 


\'M 


' 


I'li 


^liSt'i; 


u 


i      .  I 


i 


i:i 


t 


I!  hi. 


360 


THEORY  OF  STRUCTURES. 


parabola   with    its    vertex    at   A    and    having    a   parameter 


The  max.  depth  =  2CB  =  CD=  VpWJ. 

The  form  of  this  beam  is  very  similar  to  tliat  adopted  for 
cranks  and  for  the  cast-iron  beams  of  engines.  In  the  latter, 
the  material  is  usually  concentrated  in  the  flanges,  a  rib  being 
reserved  along  the  neutral  axis  for  purposes  of  connection. 

Again,  geometrical  conditions  of  transmission  require  the 
teeth  of  wheels  to  be  of  approximately  uniform  strength. 

A  cantilever  of  approximately  uniform  strength  may  be  ob- 
tained by  taking  the  tangents  CE,  DF  as  the  upper  and  lower 
edges  of  the  beam  instead  of  the  curves  CA,  DA.  The  depth 
of  the  beam  at  A  is  then  EF  =  \CD  =  ^  S'pW).  Although, 
theoretically,  the  depth  at  A  is  nil,  practicall}'  the  beam  nui.st 
have  sufficient  sectional  area  at  A  to  bear  the  shear  due  to  Jf, , 

and  the  depth  ^  V/JF,/ will  be  found  ample  for  this  purpose. 

Note. — The  dotted  lines  show  the  beams  of  uniform 
strength,  when  the  lower  edge  is  the  horizontal  line  AB. 

Ex.  2.  A  cantilever   AB   carrying  a  uniformly  distributed 

load  W,. 

At  a  distance  x  from  A, 

/=pM=^x' 


2l 


or 


y 


=  -/-^-. 


Fig,  279. 


The  beam,  in  elevation,  is  there 
fore  the  area  ACD,  AC,  AD  being  two  straight  lines,  and  the 
maximum  depth  being  • 


CD  =  2BC=l^''-^^  =  sJ 


The  sectional  area  at  A  is  nil,  as  both  the  bending  moment 
and  shear  at  that  point  are  zero. 


BEAMS  OF    UXIFOKM   STKENGTH. 


361 


Xotc. — The  (Jotted  lines  show  the  cantilever  of  uniform 
strength  when  AB  is  the  lower  edge. 

Ex.  3.  A  canlilcvcr  AB  carrying  a  zvcight  U\  at  the  free 
end  A  and  also  a  uniformly  distributed  load  l\\. 


IG.  a8o. 


At  the  distance  x  from  A, 

f  =  pM  .=  p[w,x  +  V/f^). 
Thjs  equation  may  be  written  in  the  form 


/      w  \ 


y 


pjvii 

2]]\ 


—  I. 


Theoretically,  therefore,  the  beam,  in  elevation,  is  the  area 

ACD,  the  curve  CAD  being  an  hyperbola  having  its  centre  at 

/  IV  \ 

H  i  where  AH  —  W^Ji  and  semi-axes  equal  to 


■~l    and 


The  maximum  depth  CD  =  \/ p[wj -\-  wA  =  2BC, 


«    ' 


362 


THEORY  OF  STRUCTURES. 


!■  H ; 


A  cantilever  of  approximately  uniform  strength  may  be  ob- 
tained by  taking  the  tangents  CE,  DF  as  the  upper  and  low  er 
edges  of  the  beam  instead  of  the  curves  CA^  DA.     It  may  be 

W 
easily  shown  that  the  depth  of  this  beam  at  A  is    ^     '  rr^^r), 

and  this  will  give  sufficient  sectional  area  at  A  to  bear  tlic 
shear  due  to  W^.  .    « 

Note. — The  dotted  lines  show  the  cantilever  of  uniform 
strength,  when  the  lower  edge  is  the  line  ^^.   ,  . 

Ex.  4.  A  beam  AB  supported  at  A  and  B,  and  carrying  a 

loq^d  W^  at  the  middle  point  0. 
At  a  distance  x  from  0, 


f  =  pM  =  p 


W, 


(^4 


2  \2 


Theoretically,  therefore,  tlie 

beam,  in  elevation,  is  the  area 

ACBD,  the  curves  CAD,   CBD  being  two  equal   parabolas. 

having  their  vertices  at  A   and  B,  respectively,  and  having 

parameters  equal  to  \p  \\\ . 

The  maximum  depth  =  CD  —  2CO  =  ^  V])WJ. 

A  beam  of  approximately  uniform  strength  may  be  ob- 
tained by  taking  the  tangents  CE,  CG  as  the  upper  edges 
instead  of  the  curves  CA,  CB,  and  the  tangents  DF,  DH  as 
the  lower  edges  instead  of  the  curves  DA,  DB. 

The  depth  of  the  beam  at  A  and  B  is  now  EF  =  GH  =  —  , 

2 

and  this  depth  will  give  a  sectional  area  at  the  ends  of  the  beam 

sufficient  to  bear  the  shears  at  these  point,  viz.,  — - . 

Note. — The  dotted  lines  show  the  beam  of  uniform  strength 
when  the  line  AB  is  the  lower  edge. 

Ex.  5.  A  beam  AB  supported  at  A  and  B,  and  carrying  a 
uniformly  distributed  load  W^. 


i 


!. 


BEAMS  OF   UN /FORM  STRENGTH. 


363 


At  a  distance  x  from  the 
middle  point  O, 

This  equation  may  be  writ- 
ten in  the  form 


4 


pwi 

8 


=  I. 


Theoretically,  therefore,  the  beam,  in  elevation,  is  an  ellipse 
ACBD,  having  its  centre  at  O  and  axes 


AB  =  l    and     CD 


hw^i 


The  maximum  deptii  is  of  course  the  axis  CD  ■=  2CO. 

Practically,  the  beam  must  have  a  certain  depth  at  A  and 

B  in  order  to  bear  the  shears  due  to  the  reactions  at  these 

W 
points,  viz., — ^.     If  the  horizontal  tangents  at  Cand  at  D  are 

substituted  for  the  curves,  the  volume  of  the  new  beam  is  to 
the  volume  of  the  elliptic  beam  in  the  ratio  of  4  to  rr. 

Note. — The  dotted  line  shows  the  beam  of  uniform  strength 
when  its  lower  edge  is  the  line  AB. 

Ex.  6.  A  beam  AB  supported  at  A  ayid  B,  and  carrying  a 
load  JV,  at  the  middle  point  O  and  also  a  uniformly  distributed 
load  ir,. 

At  a  distance  x  from  O, 

/  =pM  =  p\--\l-x)  +  -^'-[i  -  V)  f- 


This  equation  may  be  written  in  the  form 


-f  + 


2  IK 


-**^- 


+ 


i  j|! ,14 H\  IV,  I-  iv:  +  w-.')     g^;(4 w,  IK  +  IK'  +  rf 7) 


=  I. 


' 


I-:  '     'i'i 


:iJi 


364 


THEORY  OF  STRUCTURES. 


Theoretically,  therefore,  the 
beam,  in  elevation,  is  the  area 
ACBD,  the  curves  CAD  and 
CBD  being  the  arcs  of  ellipses 
having  the  centres  at  the  points 
K  and  L,  respectively,  where 


0K=  0L  = 


2w: 


The  maxinnum  depth  CD  =20C  =  P^  \  — ^~  -\- 


WJ  ,    WJ 


8 


}*■ 


A  beam  of  approximately  uniform  strength  maybe  obtained 
by  taking  as  the  upper  edge  the  tangents  to  the  curves  at  C. 
and  as  the  lower  edge  the  tangents  to  the  curves  at  D. 

It  may  be  easily  shown  that  the  depth  at  the  cxxdsA  and  /)' 
\y  _|_  w 
is  now  ^^— 117— i — T>7»  and  this  depth  will  make  allowance  for 

W  -\-  W 
the  shear  — ^— ' ?  at  these  points. 

Note, — The  dotted  lines  show  the  beam  of  uniform  strength 
when  the  lower  edge  is  the  line  AB. 

Case  b.  Assume  that  the  ratio  of  the  breadth  (^)  to  the 
depth  (j)  is  constant,  i.e.,  that  transverse  sections  are  similar. 


y  on  b  <x  V  M, 

or  the  ordinates  of  the  profile  of  the  beam  both  in  plan  and 
elevation  are  proportional  to  the  cube  roots  of  the  ordinates 
of  the  curve  of  bending  moments. 

For  concentrated  loads  the  bounding  curves  are  evidently 
cubical  parabolas. 

Case  c.  Assume  that  the  depth  y  is  constant.     Then 

b  ex  M, 

so  that  the  ordinates  of  the  beam  in  plan  ate  directly  propor- 
tional to  the  ordinates  of  the  curve  of  bending  moments. 
Case  d.  Assume  thf^t  th^  sectional  area  yb  is  constant. 

Then 

-     ,        ..^      ,  y  <x  M,  ';- 


na 


FLANGED    GIRDERS,  ETC, 


363 


/ 


and  the  ordinates  in  elevacion  are  directly  proportional  tu  the 
uiuiiuites  of  the  curve  of  bending  moments. 

Ill  this  beam,  the  distributioi\  of  the  material  is  very  de- 

fective,  as  the  breadth  '^  I  ~  '    /  )  must  be  infinite  when  y  —  o, 

i.e..  at  the  points  at  which  the  bending  moment  is  nil. 

Timber  beams  of  uniform  strength  are  uncommon,  as  there 
is  no  economy  in  their  use,  the  portions  removed  to  bring  the 
beam  to  the  necessary  form  being  of  no  practical  value. 

6.  Flanged  Girders,  etc. — Heams  subjected  to  forces,  of 
which  the  lines  of  action  are  at  right  angles  to  the  direction  of 
their  length,  are  usually  termed  Girders ;  a  Semi-girder,  or 
Cantilever,  is  a  girder  with  one  end  fixed  and  the  other  free. 

It  has  been  shown  that  the  stress  in  the  different  layers  of 
a  beam  increases  with  the  distance  from  the  neutral  surface,  so 
that  the  most  effective  distribution  of  the  material  is  made  by 
witlulrawing  it  from  the  neighborhood  of  the  neutral  surface 
and  concentrating  it  in  those  parts  which  are  liable  to  be  more 
severely  strained.  This  consideration  has  led  to  the  introduction 
oi  Flanged  Girders,  i.e..  girders  consisting  of  one  or  tivo  flanges 
lor  tables),  united  to  one  or  ttuo  zvebs,  and  designated  Single- 
xvebbed  or  Double-zvebbed  ( Tubular)  accordingly. 


Fig.  284. 


Fig.  a3s. 


T    T 


Fic.  a86.  Fig.  287. 


Fig.  388.  Fig.  289.  Fic.  290. 


The  web  may  be  open  like  lattice-work  (Fig.  284),  or  closed 
and  continuous  (Fig.  285). 

The  principal  sections  adopted  for  flanged  girders  are : 

The  Tee  (Figs.  286  and  287),  the  I  or  Double-tee  (Figs.  288 
and  289),  the  Tubular  or  Box  (Fig.  290). 

Classification  of  Flanged  Girders. — Generally  speaking, 
flanged  girders  may  be  divided  into  two  classes,  viz.: 


W\ 

nioR 

-  \     ' 

'>  f^','j  (   Iffin 

('■  '■''■          ■  HR 

■  -l^l   [I   \ 

'     '  -i      '  i      <     1     ' 

■     (■■      i.t    15     1    . 

^■'■'1' 

I        'h»| 

' '' iB 

V 

r 


61   »''  r 


T  i 

f 

I 

1  . 

H  ■ 

' ! ' 

t 

.■ 

•I 

v'          % 

■ 

T 


366 


THEORY  OF  STRUCTURES. 


I.  Girders  with  Horizontal  Flanges. — In  these  the  flan^'es 
can  only  convey  horizontal  stresses,  and  the  shearing  force, 
which  is  vertical,  must  be  wholly  transmitted  to  the  flanges 
through  the  medium  of  the  web. 

If  the  web  is  open,  or  lattice-work,  the  flange  stresses  are 
transmitted  throuj^h  the  lattices. 

If  the  web  is  continuous,  the  distribution  of  stress,  arising 
from  the  transmission  of  the  shearing  force,  is  indeterminate, 
and  may  lie  in  certain  curves;  but  the  stress  at  every  point  is 
resolvable  into  vertical  and  horizontal  components.  Thus,  the 
portion  of  the  web  adjoining  the  flanges  bears  a  part  of  the 
horizontal  stresses,  and  aids  the  flanges  to  an  extent  depend- 
ent upon  its  thickness. 

With  a  thin  web  this  aid  is  so  trifling  in  amount  that  it 
may  be  disregarded  without  serious  error. 

II.  Girders  with  one  or  both  Flanges  Curved. — In  these  the 
shearing  stress  is  borne  in  part  by  the  flanges,  so  that  the  web 
has  less  duty  to  perform  and  requires  a  proportionately  less 
sectional  area. 

Equilibrium  of  Flanged  Girders. — AB  is  a  girder  in  equi- 
librium under  the  action  of  external 
forces,  and  has  its  upper  flange  com- 
pressed and  its  lower  flange  cx- 
ten.ded.  Suppose  the  girder  to  be 
divided   into  two  segments  by  an 


iN 


'L. 


jL. 


B 


Fig.  291. 

imaginary  vertical  plane  MN.  Consider  the  segment  AM^'. 
It  is  kept  in  equilibrium  by  the  external  forces  on  the  left  of 
jSIN,  by  the  compressive  flange  stress  at  iV  (  =  C),  by  the 
tensile  flange  stress  at  J/  (  =  T),  and  by  the  vertical  and 
horizontal  web  stresses  along  MN.  The  horizontal  web 
stresses  may  be  neglected  if  the  web  is  thin,  while  the  vertical 
web  stresses  pass  through  M  and  jV,  and  consequently  have  no 
moments  about  these  points. 

Let  d  be  the  effective  depth  of  the  girder,  i.e.,  the  distance 
between  the  points  of  application  of  the  resultant  flange  stresses 
in  the  plane  MN.  ,.,  ^ 

,_     Take  moments  about  J/ and //successively.     Then 

Cd  =  the  algebraic  sum  of  the  moments  about  M  of 


ssscs  are 


FLANGED   GIRDERS,  ETC. 


367 


the  external  forces  upon  AMN  —\}ci^  bending  moment  at 
Mis'  =  M. 

So,  Td  =  M;  .'.  Cd=M^  Td,  and  C  =  T. 

Hence,  the  flange  stresses  at  anv  vertical  section  of  a  girder 
are  equal  in  magnitude  but  opposite  in  kind.  The  flange 
stress,  whether  compressive  or  tensile,  will  be  denoted  by  F. 

Example. — A  flanged  girder,  of  which  the  effective  depth 
is  10  ft.,  rests  upon  two  supports  80  ft.  apart,  and  carries  a  uni- 
formly distributed  load  of  25CK)  lbs.  per  lineal  foot.  Determine 
the  flange  stress  at  10  ft.  from  the  end,  and  find  the  area  of 
the  flange  at  this  point,  so  that  the  unit  stress  in  the  metal 
may  not  exceed  10,000  lbs.  per  square  inch. 

The  vertical  reaction  at  each  support 


80  X  2500 


{1  wKk 
111 

■i. 

1 

i  '^ 

,!' 

i  1 

1 

H 

,1'   )'• 


=  100,000  lbs. 


.-.  F.\o  =  M=  looooo  X  10  —  2500  X  10  X  5  =  875,000  ft.-lbs. 

.•.F=  87,500  lbs. 


!:i 


in. 


^,  .     ,  87500 

The  required  area  = =  8.75  sq.  1 

^  lOOOO  ^ 

Cor.  I.  Fd=M=^I=-^^I. 
K         y 

Cor.  2.  At  any  vertical  section  of  a  girder, 
let  <?, ,  a, ,  be  the  sectional  areas  of  the  lower  and  upper  flanges, 
respectively ; 
/,,/,,  be  the  unit  stresses  in  the  lower  and  upper  flanges, 
respectively.     Then 

and  the  sectional  areas  are  inversely  proportional  to  the  unit 
stresses.    ,. 

This  assumes  that  F  is  uniformly  distributed  over  the 
areas  <r, ,  a, ,  so  that  the  effective  depth  is  the  vertical  distance 
between  centres  of  gravity  of  these  areas.  Thus,  the  flange 
stresses  at  the  centres  of  gravity  are  taken  to  be  equal  to  the 


368 


THEORY  OF  STRUCTURES. 


maximum  stresses,  and  the  resistance  offered  by  the  web  to 
bending  is  disregarded.  The  error  due  to  the  former  may 
become  of  importance,  and  it  may  be  found  advisable  to  make 
the  effective  depth  a  geometric  mean  between  the  depths  from 
outside  to  outside  and  from  inside  to  inside  of  the  flanges. 

Ti:us,  if  these  latter  depths  are  /r, ,  //, ,  the  effective  depth 
=  i/p,  (Art.  7).  '•  •  • 

Example  i.  At  a  given  vertical  section  of  a  flanged  girder 
the  sectional  area  of  the  top  flange  is  lO  sq.  in.,  and  the  cor- 
responding unit  stress  is  8000  lbs.  per  square  inch.  Find  the 
sectional  area  of  the  lower  flange,  so  that  the  unit  stress  in  it 
may  not  exceed  10,000  lbs.  per  square  inch, 
a, .  lOOCD  =■  F  —  10 .  8000  ;     .'.  rt,  =  8  sq.  in.  and  F  =  80,000  lbs, 

Ex.  2.  A  wrought-iron  girder  weighing  iv  lbs.  per  lineal 
ft.,  of  /  ft.  span  and  d  ft.  depth,  has  horizontal  flanges  and 
a  uniform  cross-section.  The  weight  of  the  web  is  rqual  to  the 
weight  of  the  flanges.  Show  that  if  the  coefficient  of  strength 
is  9000  lbs.  per  square  inch,  the  limiting  value  of  /  is  5400X'  ft., 
k  being  the  ratio  of  depth  to  span. 

Maximum  flange  stress  =  -^- ;         ,' 


Area  of  each  flange        = 


W 


9000 .  8^/ 


m. ; 


^1;S    ^^1 


and 


Total  sectional  area        = -r-,  in., 

9000 .  M 

4w/' .  / 


total  volume  of  girder  in  feet  = 


Hence, 


and 


zv/  =  total  weight  = 


9000.  Sd.  144 

4«'/'.48o 
9000.  8d.  144'' 


/  =  5400^-  =  5400/t. 


I  \ii 


FLANGED   GIRDERU,  ETC. 


369 


Note. — The  compressive  strength  of  cast-iron  is  almost  six 
times  as  great  as  tiic  tensile  strength,  and  therefore  the  area 
of  tlie  tension  flange  of  a  girder  of  this  material  should  be 
about  six  times  that  of  the  compression  flange.  Considering, 
iioucver,  the  difficulty  there  is  in  obtaining  sound  castings, 
ami  also  the  necessity  to  provide  sufficient  lateral  strength,  il 
b\  no  means  follows,  nor  is  it  even  probable,  that  the  ratio  of 
ultimate  strengths  is  the  best  for  the  working  strengths.  Some 
aiitiiorities  are  of  the  opinion  that  girders  should  be  designed 
with  a  view  to  their  elastic  strength,  and  that  therefore  the 
working  unit  stresses  in  the  case  of  wrought-iron  and  steel 
sluiuld  be  equal,  if  this  will  insure  sufificient  lateral  stability, 
ami  in  the  rati(j  of  2  to  1  or  3  to  '  for  cast-iron,  which  will  give 
sufficient  lateral  stability  and  niake  allowance  for  defective 
castings. 

rhe  formula  W '—  Cy  is  often  employed  to  determine  the 

strength  of  a  cast-  or  wrought-iron  girder  which  rests  upon  two 
supports  /  inches  apart,  d  being  i.s  depth  in  inches,  and  a  the 
net  sectional  area  of  the  bottom  flange  in  square  inches.  C  is 
a  constant  to  be  determined  by  experiment.  Its  average  value 
for  cast-iron  is  24  or  26,  according  as  the  girder  is  cast  on  its 
side  or  with  its  bottom  flange  upwards.  An  average  value  of 
C  for  ivroii^i^/it  iron  is  80. 

Cor.  3.  A  girder  with  horizontal  flanges,  of  length  /  and 
depth  d,  rests  upon  two  supports,  and  is  uniformly  loaded  with 
a  weight  iv  per  unit  of  length. 

The  bending  moment  at  a  vertical  plane  distant  x  from  the 
centre  is  • 

..=^(|..)_.(i_.)i(i_.).-(,_ir). 

Aho,  M  =  Fd  =  a/d,  a  being  the  sectional  area  of  either 
tlairjje  at  the  plane  under  consideration,  and  /  the  correspond- 
ing unit  stress. 


^^      tt'/V        4x'\ 


y 


":  'm 


370  THEORY  OF  STRUCTURES. 

Let  A  be  the  flange  sectional  area  at  the  centre.     Then 


Afd- 


8 


Hence 


.  =  4-^'). 


an  expression  from  which  the  flange  sectional  area  at  any  point 
of  the  girder  may  be  obtained  when  the  area  at  the  centre  is 
known. 

Cor.  4.  F  represents  indifferently  the  sum  of  the  horizontal 
elastic  forces  either  above  or  below  the  neutral  axis,  and  is 
therefore  proportional  to  A,  the  sectional  area  of  the  girder; 
d  is  the  distance  between  the  centres  of  resultant  stress  and  is 
proportional  to  D,  the  depth  of  the  girder. 

.-.  M  <xAD=  CAD, 

a  form  frequently  adopted  for  solid  rectangular  or  round  gird- 
ers, but  also  applicable  to  other  forms. 

Remark. — The  effective  length  of  a  girder  may  be  taken  to 
be  the  distance  from  centre  to  centre  of  bearings. 

The  effective  depth  depends  in  part  upon  the  character  of 
the  web,  but  in  the  calculation  of  flange  stresses  the  following 
approximate  rules  are  sufficiently  accurate  for  practical  pur- 
poses : 

If  the  web  is  continuous  and  very  thin,  the  effective  depth 
is  the  full  depth  of  the  girder. 

If  the  web  is  continuous  and  too  thick  to  be  neglected,  the 
effective  depth  is  the  distance  between  the  inner  surfaces  of 
the  flanges. 

If  the  web  is  open  or  lattice-work,  the  effective  depth  is  the 
vertical  distance  between  the  points  of  attachment  of  the 
lattices. 

If  the  flanges  arc  cellular,  the  effective  depth  is  the  distance 
between  the  centres  of  the  upper  and  lower  cells. 


m 


mm 


hen 


.ny  point 
centre  is 

orizontal 
is,  and  is 
e  girder ; 
ess  and  is 


,und  gird- 
taken  to 

racter  of 
following 
cal  pur- 


Iti 


live 


depth 


jcted, the 
jrfaces  of 

)th  is  the 

of  the 

distance 


EXAMPLES  OF  MOMENTS  OF  INERTIA. 


371 


A 

Fig.  39». 


7.  Examples  of  Moments  of  Inertia. — (rf)  Doiible-tce  Sec- 
tion.— First,  suppose  the  web  to  be  so  thin  that 
it  may  be  disregarded  without  sensible  error. 

Let  the  neutral  axis  pass  through  G,  the  cen-    -I 
tre  of  gravity  of  the  section.  f' 

Let  rt, ,  rt:,  be  the  sectional  areas  of  the  lower 
;ind  upper  flanges,  respectively,  and  assume  that 
each  flange  is  concentrated  at  its  centre  line. 

Let  /<!, ,  //,  be  the  distances  of  these  centre  lines  from  G. 

Let  h,  +  /r,  =■  d. 

Approximately,  /  =  aji^  -(-  aji^.  .         . 

Also,      {a^  -\-  a^h^  =  a^d,     and     («,  +  a,)//,  =  a^d. 

Again,  if  /, ,  /,  are,  respectively,  the  unit  stresses  in  the 
metal  of  the  lower  and  upper  flanges, 

/.  / 

M  =  J- 1  =  ffi^d,    and  also     =  j-V  =  f^a^d. 

If  rt,  =  rt,  =  a,  /,  —f^—f,  suppose,  and  M  =.  fad. 
Second.  Let  the  web  be  too  thick  to  be  neglected. 
As  before,  let  the  neutral  axis  pass  through  G,  the  centre 
^_____^  of  gravity  of  the  section. 

Let  rt, ,  rt,  be  the  sectional  areas  of  the  lower 
and   upper  flanges,  respectively,  and  assume  that 


-K- 

I 


f  each  flange  is  concentrated  at  its  centre  line. 
Fig.  292a.  j^g^  ^^  ^  ^^  j^g  ^j^g  sectional  areas  of  the  portions 

of  the  web  below  and  above  G^  respectively. 

Let  //, ,  /r,  be  the  distances  from  G  of  the  lower  and  upper 
flange  centre  lines.  •         .    <  .  '■ 

Let  h,  -f-  //,  =  d. 

Approximately, 

.  =  .,,-  +  ..(^  +  f)  +  ...-  +  .(^+¥)       :    , 


I 


^ll 


.1  ■ 


,  n  r 


^^:^"^ 


f^\ 


172 


THEORY  OF  STRUCTURES. 


Also,  [a^  +  ~)^i  =  (^a  +  ~Yhi  and  this  equation,  together 

with  h^  -\-  h^  =  d,  will  give  the  values  of  /i^ ,  h^ ;    hence  the 
value  of  /  may  be  determined. 


As  before,  7/=  J/ =^V. 


A' 


Let  a^=  a^  =  A  and  a^=^  a^=l  -. — .    Then 


V.'\  li'i' 


M 


ill*!    '   'i 


Hence, 


d  I  A'\d'' 

K  =  K  =  ~,     and     /=(^^  +  --j--. 


M=%[A^^)^^^f[A^^Y 


/"being  the  unit  stress  in  either  flange. 

Thus,  the  web  aids  the  girder  to  an  extent  equivalent  to  the 
increase  which  would  be  derived  by  adding  one-sixth  of  the 
web  area  to  each  flange.  -  If  the  weight  of  the  material  remains 
constant,  M  increases  with  d.  At  the  same  time  the  thickness 
of  the  web  diminishes,  its  minimum  value  being  limited  by  cer- 
tain practical  considerations  (Art.  8).  Hence  it  follows  that 
the  distribution  of  material  is  most  effective  when  it  is  concen- 
trated as  far  as  possible  from  the  neutral  axis  (Art.  5). 

N.B. — It  must  be  remembered  that  /,  and  _/,  are  not  the 
maximum  stresses.  If  /, ,  /,  are  the  thicknesses  of  the  lower 
and  upper  flanges,  respectively,  then 


maximum  tension 


and 


=  /. 


K  +  K 

h,      ' 


maximum  compression  =/, — -. . 


EXAMPLES   OF  MOMENTS  OF  INERTIA. 

Again,  take  moments  about  G.    Then 


373 


or 


aJ,^A'^-^—^=aJ,, 


N  : 


W  ' 


i    (I 


which  gives  a  relation  between  the  flange  and  web  areas  if 
/, ,/,  are  known. 

For  example,  take/",  =  24/,.    Then 

a,-\A'  =  %a^, 

a  formula  which  agrees  very  closely  with  modern  practice  in 
cast-iron  girders. 

If/,  =/,,  a,  =^,. 

The  principles  of  construction  require  a  beam  or  girder 
to  be  designed  in  such  a  manner  as  to  be  of  uniform 
strength,  i.e.,  equally  strained  at  every  point.  An  exception, 
however,  is  usually  made  in  the  case  of  timber  beams  or  girders. 
The  fibres  of  this  material  are  real  fibres  and  ofYer  the  most 
effective  resistance  in  the  direction  of  their  length,  so  that  if 
they  are  cut,  their  remaining  strength  is  due  only  to  cohesion 
with  the  surrounding  material.  Besides,  there  is  no  economy 
to  be  gained  by  removing  a  lateral  portion,  as  the  waste  is  of 
little,  if  any,  practical  value. 

Example.  The  lower  and  upper  flanges  of  the  section  of  a 
girder  are  I  in.  and  i^  in.  thick,  respectively,  and  are  each  24  in. 
wide  ;  the  effective  depth  of  the  girder  is  48  in.,  and  the  web 
is  I  in.  thick.  Determine  the  position  of  the  neutral  axis  ;  also 
find  the  flange  unit  stresses  when  the  bending  moment  at  the 
given  section  is  250  ft.-tons.     Using  the  preceding  notation, 

rt!,  =  24  sq.  in.,    a,  =  36  sq.  in.,    and    «,  -f-  ^^  =  24  sq.  in. 

The  centre  of  gravity  of  the  web  is  half-way  between  AB 
and  CD.    Thus, 

'      24^.  -f  24(/r,  -  24)  =  36(48  -  /^), 


V%. 


i 


IM 


H  I' 


374 
or 


rilEORY  OF  ST/iUCTU/iES. 


IQ2  144 

/i,  =  -^~     and    //,  =        - ,  defining  the  position  of  G. 

.■-..;.',',?.  *'  ■  '•  ' 

Again, 

IQ2    I       96        .  ,  144    I       72 

a.  =  -^- .  —  =  —  sq.  in.      and     a.  = .  -  =  --  sq.  in. 

'727  727 

'"":•. /=a4-ff)(f)'+(36+^)(f)=^5f^. 

Also, 

M  —  250  ft.-tons  =  3000  inch-tons. 

7        267264  7     ^  267264 

192-''       7  -^  144  7  -"    J^        yi      :) 

.  •    .•.  /,  =  2/^  tons  per  sq.  in.     and     /,  =  ii^|  tons  per  sq.  in. 

Third.  It  is  often  convenient  to  calculate  the  moment  of 
inertia  of  a  built  beam  symmetrical  with  respect  to  the  neutral 
axis,  as  follows : 

Let  Fig.  293  represent  the  section  of  such  a  beam,  com- 
posed of  equal  flanges  connected  with  the  web  by  four  equal 
angle-irons. 

_       Let  the  width  AFoi  the  flange  =  a. 
'n        "     the  side  BC{=  BE)  of  an  angle-iron  =  k 
"     thickness  GN{=  KL) of  an  angle-iron  =/. 
"    MN=DE-KL  =  h-f=c. 

AF—BE      a-{2b-\-c) 


aC^ 


tK' 


^ 


EF  = 


=  d. 


"     k^  be  the  outside  depth  of  the  section. 
"     h^  "     "     depth  between  flanges. 


Let  /r,  be  the  depth  between  the  faces  MN,  M'N' 
"     h,  "     "         "  "  •'       "      KL,  K'L'. 


/  = 


ahl 
12 


2\^w/^:  +  ch:+/h:)\, 


EXAMPLES  OF  MOMENTS  OF  INERTIA. 


375 


In  this  value  of  /,  the  weakening  effect  due  to  the  rivet- 
holes  in  the  tension  flange  has  been  disregarded.  If  it  is  to  be 
taken  into  account,  let/  be  the  diameter  of  the  rivets. 

The  centre  of  gravity  of  the  section  is  now  moved  towards 
the  compression  flange  from  its  original  position  through  a 
distance 

and  the  moment  of  inertia  of  the  net  section  with  respect  to 
the  axis  through  the  new  C  of  G.  is 


•G, 


A'  being  the  net  area  of  the  section,  and  /  having  the  value 
given  above. 

Fourth.  The  value  of  /  for  a  double-tee  section   may  be 
more  accurately  determined  as  follows  : 

Let  the  area  of  the  top  flange  be  yi, ,  and 
its  depth  //,. 

Let  the  area  of  the  bottom  flange  be  ^,, 
and  its  depth  /;,. 

Let  the  area  of  the  web  flange  be  A^ ,  and 
its  depth  //,. 

Let  yi,-(-  A^-\-  A^=  A,  and  /t,-{-  /(,-f-  /i^=/i.  Fig.  294. 

Let  G   be  the  centre  of  gravity  of  the  section. 
"     G,      "  "  "  "      top  flange. 

"     G,      "  "  "  "      web. 

"     G,      "  "  "  "      bottom  of  flange. 

Let  j\  be  the   distance  of  C  from   the   upper  edge  of  the 
section. 

Let  y,  be  the  distance  of  G  from  the  lower  edge  of  the 
section.  •       '     .  :  "   '  .  .  *    -  - 

Take  moments  about  G,.     Then 


•G., 


{A,-\.A,  +  A,)GG,  =  A,.G,G,-\-A,.G,G, 

f/t,    .    .     .   K 


=  .^(^,,.,^.)^,,(|+|), 


I.",-'' 


\ 


:,.r,y 


-.      -t 


3;,i: 


lii 


or 

So, 
and 
Hence, 


THEORY  OF  STRUCTURES. 


^.(>'^.  +  2/',  +  /0  +  A(/^  +  /0 


yjyj^  - 

2A 

GG,-- 

-A,{A,+/i,)  +  A,{/t,  +  /i,) 

2A 

nn  - 

A,{/i,  +  /i,)  +  AX/i,-\-2/i,  +  /i,) 

2A 


//,    A,{h,+7ji,\-h:)+A,{K+h:) ,  h   //,    /., 

^,=  6^6^3+7  = ^ -^-i-~2-~2 

_h        (/^,+/,.)(^,+yj,4-^.)-^,(//.4-2//,+//.)-/JA+/0 
2  2A 

h  __  AlK  +  /Q  -  ^.(/''.  +  /Q  -  ^IK  -  ^)     ' 

~   2  2A 


So, 


y,  =  GG,  +  ^-  =  etc 


Again,  /,  with  respect  to  G, 

=  :fl2!Lj^A,.G,G'-^^^^^  +  A,.G,G'-^A,^+A,.G,G' 

12        '         '        '  '        12        '        '  '  12     ' 

AA'  +  AA'  +  AJi,*  ' 


/,  being  equal  to 
Hence, 


A 


/  =  /.  +  ^  i  ^,(/^  +  ^)  +  ^,(/^  H-  2/^  +  /^) }' 

+  ^]^  A  +  2//,  +  A,)  +  ^,(>^,  +  /l,)\\ 


EXAMPLES  OF  MOMENTS  OF  INERTIA. 


m 


=^-+:M-i 


=  /.+ 


4^' 


=  /,4- 


=  /, 


4^' 


+  A,A,\h,  +  A.)'  +  A,A:{h,  +  2//.  +  V 

4- (^3^^' +  ^sM.XA  +  2//,  + /O' 

L+  2^,^,^3(7/,  4-  2/r,  +  f^.W^,  +  Z-',  +  /^,  +  /O 

/^,.>i3(//.  +  //,)M  +  A,Alh,  +  /^M  ] 
-A,A,Alh,-^2K  +  h,r 
+  (/J./i,'  +  ^3M,)(//.  +  2/.,  +  /0' 
+  2^A^s(/';.  +  2//,  +  /0' 


Hence,  finally, 

AX^AX  +  AA 


I- 


12 


4-  ^\A,Alh,  4-  //;/  +  A^Alh,  4-  h,r  4-  ^,^3(/..  4-  2//,  4-  h,y\. 

Cor.  I.  If  //,  and  /*,  are  small  compared  with  //,,  put 


/:,-  =  h' 


Then 


y  = 


AM'+A^{/.'  +  '^)      ,^      ,, 


2A 


2A,  +  A, 

-^ ,  nearly, 


m'-^m 


;ii: 


i( 


i 'i 


'I 

1 


I 


< 


Vr. 


37« 
and 


/  = 


THEORY  OF  STRUCTURES. 


/f///  +  A(//'-'^^^')'  +  /M.' 


12 


(   12   "^  ~4//  ) 

Note. — If  A^  is  also  very  small,  as  in  the  case  of  an  open 
web,  then 

A  A  A 

j'^  =  /{'!-     and     /= /<'"—-—-',  approximately. 

Cor.  2.  Let  )',,.. I'a  be  the  distances  of  G  from  the  uppei  aiul 
lower  edges,  respectively ;  let  /], ,  //,  be  the  corresponding 
maximum  working;  unit  stresses. 

From  the  preceding  corollary, /*  =  j,  =  — '~T~~'' 


or 


^.  +  A  +  ^.    y.^-yk 


2.-i,  +  W. 


2^» 


Jb  ^j/b  Jb  ^Jb 


Hence, 


/. 


^.  +  tV  =  ^, 


EXAMl'LES  Oh    MOMENTS  OF  INENTIA. 


%n 


I  12  "•"  4A        ,        .  }       ' 

m   ■ 

_j^,i  4A,A,  +  a:  f-  ^,A,  4-  \2A,A,  \   ' 
\  12A  ) 


12 


12 


yl 


^•'(^/:") + '^■^'("  TT^I + 7;(^^.^.+ 12/^.') 


Lij:ltt2A  4-  yi  1 


J 


Fifth.  T-section. 

Let  the  area  of  the  flange  be  /J, ,  and  its 
depth  //,. 

Let  the  area  of  the  web  be  A^ ,  and  its 
depth  /;,. 

Let  A,'^-  A,  =  A,     and     h,-\-h^  =  h. 

Let  G  be  the  centre  of  gravity  of  the  sec- 
tion, G^  of  the  flange,  and  C,  of  the  web. 


G 
8, 


Flo.  295. 


Let  J,  be  the  distance  of  G  from  foot  of  the  web. 
Then 


(/t.H-w,)(//.+//,)  ^  ax- A  A 


U   ! 


u- 


und 


TJ/HOA'V  OF  S/A'UCrUAH^i. 


•^'  2        "^    2(//,-f /i.)  2    "^  2^ 


^.(;=-'  4- //.->'.  =  ^3^      and     ^,^=;..--'=_L-. 
Hence  /,  with  respect  to  a  liorizontal  line  through  G^ 

whicli  reduces  to 

Ajt:  +  Ajt:    A.Aji^ 

Cor.  I.  If  /l^  is  very  small  as  compared  with  //, ,  put 


then 


}\{A,+A:)  =  AJi'  +  /J,(^^'-  -  y  =  [a,  -f  ^-')  A',  nearly, 


or 


2K       A 


)■ 


and 


AA'  +  A,{/,'-^^)'       A,A,[/i'  +  ^^J 


12 


44 


or 


\I2  4^    / 

Cor.  2.  Let  j'a  be  the  distance  of  the  compressed,  or  upper, 
side  from  the  neutral  axis. 


TO    DUSIGN  A    UIKDKK   VF    UNll-ONM   STRENGTH.       38 1 

Let  J'/,  be  tlic  distance  of  the  stretched,  or  lower,  side  from 
tlic  neutral  axis. 

Lct/a  be  the  crushing  unit  stress, /» the  tensile  unit  stress. 

I'rom  the  preceding,  y^  —  '  "^  '  J  — -' ;    but  h'  =7,.  -\-y^ ; 

2  A 


2     /1. 4-^', 


J.,  =='------—, J,    and    A,=:A/~'^-'^=:A/'     ^' 


2jft  •      2// 


IIciicc,  /  becomes 


_  /r        2/.  -/,                   /.  +/,_y^  +  y,  _  //' 
_       /I      -  ^     anu     — - — . 

AWr. — Although  the  preceding  approximate  methods  arc 
often  useful,  they  can  only  be  regarded  as  tentative  and  shoukl 
always  be  checked  by  an  accurate  determination  of  the  moment 
of  inertia  and  of  the  position  of  the  neutral  axis. 

8.  To  design  a  Girder  of  Uniform  Strength,  of  an 
I-section  with  equal  Flange  Areas,  to  carry  a  Given 
Load. 

Let  J  be  the  depth  of  the  girder  at  a  distance  x  from  its 
middle  point. 

Let  A  be  the  sectional  area  of  each  flange  at  a  distance  x 
from  its  middle  point. 

Let  A'  be  the  sectional  area  of  the  web  at  a  distance  x  from 
its  middle  point. 

Lot  M  be  the  bending  moment  at  a  distance  x  from  its  mid 
tile  point. 

Let  5  be  the  shearing  force  at  a  distance  x  from  its  middle 
point.    Then     .  ,,     • 


f(A+'~)y  =  M, 


/being  the  safe  unit  stress  in  tension  or  compression. 


:i.; 


382 


THEORY  OF  STRUCTUKES. 


Web. — Assume  tli.it  the  web  transmits  the  xv/toic  of  tlic 
shearing  force.  This  is  not  strictly  correct  if  the  flanj;;^  is 
curved,  as  tlie  flange  then  bears  a  portion  of  the  shearing  force. 
The  error,  however,  is  on  tlie  safe  side. 

Theoretically,  the  web  should  contain  no  more  material  than 
is  absolutely  necessary. 

Let/,  be  the  safe  unit  stress  in  shear.     Then 


/J'  = 


I: 


/.' 


and  the  sectional  area  is,  therefore,  independent  of  the  depth. 


The  tJiickncss  of  the  web  = 


A' 


5 

7./ 


but  this  is  often  too  small  to  be  of  any  practical  use. 

Experience  indicates  that  the  minimum  thickness  of  a  plate 
which  has  to  stand  ordinary  wear  and  tear  is  about  \  or  -^^  in., 
while  if  subjected  to  saline  influence  its  thickness  should  be 
I  or  ^  in.  Thus,  the  weight  of  the  web  rapidly  increases  with 
the  depth,  and  the  greatest  economy  will  be  realized  for  a  cer- 
tain definite  ratio  of  the  depth  to  the  span. 

The  thickness  of  the  web  in  a  cast-iron  girder  usually 
varies  from  i  to  2  in. 

In  the  case  of  riveted  girders  with  plate  webs  of  medium 
size,  all  practical  requirements  are  effectively  met  by  specifyiiiif 
that  the  shearing  stress  is  not  to  exceed  one-half  o{  the  flaiii,fe 
tensile  stress,  and  that  stiffeners  are  to  be  introduced  at  inter- 
vals not  exceeding  txvice  the  depth  of  the  girder  when  tlie 
thickness  of  the  web  is  less  than  one-eightieth  of  the  depth. 
Again,  it  is  a  common  practical  rule  to  stiffen  the  web  of  a 
plate  girder  at  intervals  approximately  equal  to  the  depth  of 
the  girder,  whenever  the  shearing  stress  in  pounds  per  square 

inch  exceeds  12000-^  (i  -\ ),  H  being  the  ratio  of  the 

depth  of  the  web  to  its  thickness.      •  , '  ■ 

Flanges. — First.  Assume  that  the  flanges  have  the  same 
sectional  area  from  end  to  end  of  girder. 


TO  DESIGN  A    GIKDEK   OF   UN  1 1- ON  M   STRENGTH.      383 

If  the  effect  of  the  vvcb  is  nef^lected, 

_   M 
^  ~ /A' 

and  the  depth  of  the  beam  at  any  point  is  proportional  to  the 
ordinate  of  the  bendin^-moinent  curve  at  the  same  point. 

l""or  example,  let  the  U)ad   be  uniformly  distributed  and  of 
intensity  w\  and  let  /  l)e  the  span.     Then 


M 


=  2(4-4 


o 

Kn..  ay6. 


and  the  beam  in  elevation  is  the  parabola  ACH,  havinj.'   its 
vertex  at  C  and  a  central  depth  CO  =  i.   i">     The  depths  thus 

determined  are  a  little  greater  than  the  depths  more  correctly 
given  by  the  equation  • 

M 


/(''+^:-) 


Second.  Assume  that  the  depth  j  of  the  girder  is  constant. 
Then 

and,  neglecting  the  effect  of  the  web,  the  area  of  the  flange  at 
any  point  is  proportional  to  the  ordinate  of  the  curve  of  bend- 
ini,^  moments  at  the  same  point. 

Let  the  load  be  uniformly  distributed  and  of  intensity  w, 
also,  let  the  flange  be  of  the  same  uniform  width  ^  throughout. 


gjLc 

C 

ff 

=4. 

A  1    2 

3 

0 

S     S 

1   B 

Fiii.  j'iT. 

The  flange,  in  elevation,  is  then   the  parabola  ACB,  having 

its  vertex  at  C  and  its  central  thickness  CO  =  xtt.-      Such 

8/yo 


w:% 


II 


-.LiM 


384 


THEORY  OF  STRUCTURES. 


beams  are  usually  of  wrought-iron  or  steel,  and  are  built  up  hy 
means  of  plates.  It  is  impracticable  to  cut  these  plates  in  sucii 
a  manner  as  to  make  the  curved  boundary  of  the  flange  a  true 
parabola  (or  any  other  curve).  Hence,  the  flange  is  generally 
constructed  as  follows: 

Draw  the  curve  of  bending  moments  to  any  given  scale. 
By  altering  the  scale,  the  ordinates  of  the  same  curve  will 
represent  the  flange  thicknesses.  Divide  the  span  into  seg- 
ments  of  suitable  lengths. 

From  A   to  i    and  B  to  7  the  thickness  of  the  flange  is 

\a  =  7/;  from  1  to  2  and  7  to  6  the  thickness  is  2b  =^  6e;  from 

2  to  3  and  6  to  5  the  thickness  is  y  =  ^d\  and  from  3  to  5  the 

thickness  is  CO. 

M      A'\    . 
: -2' I    IS   somewhat 

less  than  that  now  determined,  but  the  error  is  on  the  safe 
side. 

Again,  at  any  section, 


The   more   correct  value  of  A 


E 
R 


2f 


and  hence      R  oc  y,  the  depth. 


Thus  the  curvature  diminishes  as  the  depth  increases,  so 
that  a  girder  with  horizontal  flanges  is  superior  in  point  of 
stiffness  to  one  of  the  parabolic  form.  The  amount  of  mct.il 
in  the  web  of  the  latter  is  much  less  than  in  that  of  the 
former.  If  great  flexibili!:y  is  requircil,  as  in  certain  dyna- 
mometers, the  parabolic  form  is  of  course  the  best. 

9.  Deflection  of  Girders.--The  principles  of  economy  and 
strength  require  a  girder  to  be  designed  in  such  a  manner  that 
every  part  of  it  is  proportioned  to  the  greatest  stress  to  which 
it  rnay  be  subjected.  When  such  a  girder  is  acted  upon  by 
external  forces,  it  is  uniformly  strained  throughout,  and  in 
benuing,  the  neutral  axis  must  necessarily  assume  the  form  nt 
an  arc  of  a  circle,  provided  the  limit  of  elasticity  is  not  ex- 
ceeded. It  might  be  supposed  that  the  curve  of  deflection  is 
dependent  upon  the  character  of  the  web,  and  this  is  doubtless 
the  case,  but  experiments  indicate  that  so  long  as  the  flani;e 


iM ,  liM  i  I 


uilt  up  by 

es  in  svicii 

:loe  a  true 

generally 

iven  scale. 

curve  will 

into  seg- 

e  flange  is 
=  6^ ;  from 
I  3  to  5  the 

somewhat 
3n  the  sate 


ncreases,  so 

in  point  of 

t  of  mct.il 

hat  of  the 

tain   dyna- 


n 


onomy  ^mJ 
anner  th;it 

ss  to  which 
d  upon  by 
lut,  and   ill 

the  form  "t 

is  not  ex- 

eflection  is 

s  doubtle -.s 

the  flange 


DEFLECTION  OF  GIRDERS. 


38; 


unit  stresses  are  unaltered  in  amount,  the  influence  of  the  web 
ma}'  be  disregarded  witliout  sensible  error. 

Let y" be  the  unit  stress  in  the  beam  at  a  distance  y  from 
the  neutral  axis;  let  d  be  the  depth  of  the  beam.     Then 

/      M      E 

-  =   y  =  ^  =  a  constant, 

y        I        K 

assuming  that  the  neutral  axis  is  an  arc  of  a  circle  of  radius  R. 


But    y  a  d,     and 


I  =  Ak'  o^  Ad\ 


Hence/a  jv  a  d;  and  if  the  depth  is  constant,  /  is  also  con- 
stant and  the  beam  is  of  uniform  strength. 
If  the  area  A  is  constant, 

d  a  ^/ j7. 

Example   i.    A  cantilever  bent  under  the  action  of  exter- 
nal forces,  so  that  its  neutral  axis  AB  assumes 
the  form  of  an  arc  of  a  circle  having  its  centre  ^E^^^!r\~f 
at  0.  pl""*^ 

Draw  the  verticals  OA,  BF,  and  ihe  horizon-     j  '  ^ 

tals  BR,  FA.  \ 

The  vertical  deviation  of  B  from   the  hori-     | 
zontal,   viz.,  BF,   is    the    maximum    deflection. 
Denote  it  by  D.  0 

Let  radius  of  circle  =  R.  '"^'  "^^^ 

Since   the   deflection   is  very  small,  BE  is   approximately 
equal  to  AB  (  =  /),  the  length  of  the  cantilever. 

.-.  /-■  =  BE'  =  AE{2R  -AE)=  iRD  ~  IT  =  2RD, 

a.s  17  may  be  disregarded  without  much  error. 

Also,  the  deflection  at  any  point  distant  x  from  A   is  evi- 

"J^'itly  :^.     If /is  the  stress  in  the  material  at  a  distance  y 

from  the  neutral  axis, 

/      E       2DE  ^      7DEy 

v  =  R  =  --7^     °''    ^=-^- 


v 


ii  <"  ii . 

.1! 


,'  . 


' !  '.      •  -■■ 

% 

*■ 

i 

1 

■  i  i 

■ 

''^il 

■m 

1|: 


386 


THEORY  OF  STRUCTURES. 


if  M  X\ 


Ex.  2.  A  girder   resting    upon  two  supports  at  A   and  B 

is  bent  under  the  action  of  external 
forces  so  that  its  neutral  axis  ACB 
assumes  the  form  of  an  arc  of  a  circle 
having  its  centre  at  O. 

Draw  the  vertical  OC,  meeting  the 
vB  horizontal  AB  in  F. 

CF  is   the   maximum   deflection ; 
denote  it  by  D. 

Since  D  is  very  small,  its  square 
may  be  disregarded  and  the  horizontal  AB  may  be  supposed 
equal  to  the  length  ACB  {  =  /)  of  the  girder,  without  much 
error.     Then 

^'      AF'  =  FC{2R  -  FC)  =  2RD  -  D' =  2RD. 


Fig.  299. 


4 
Hence, 


A,  •  /       ^ 

Also,  smce  —  =  ^, 

y      K 


D  = 


SR' 


/  = 


%DEy 


jj 


The  deflection  at  a  distance  x  from  F  =.  D ^. 

2R 

Ex.  3.  A  timber  beam  of  20  ft.  span,  is  12  in.  deep  and  6 
in.  wide :  what  uniformly  distributed  load  ( W)  will  deflect  the 
beam  i  in.,  E  being  1,200,000  lbs.? 

By  Ex.  2, 


• 

Also, 

.  I<  =  7200  m. 

W.20            E^ 
S--''-R^- 

1200000  d(^'       1200000  6.  12' 
7200   12    7200    12 

.•. 

W  =  4800  lbs. 

nwrfmrr 


1 


DEFLECTION  OF  GIRDERS. 


387 


and  B 
jxternal 

is  Acn 

a  circle 

:ting  the 

;flection ; 

ts  square 
supposed 
3ut  much 

ID. 


Ex.  4.  Let  s^,  f^,  d^,  and  s^,  f^,  d^,  respectively,  be  the 
length,  unit  stress,  and  distance  fronti  the  neutral  axis  of  the 
stretched  and  compressed  outside  fibres  in  Examples  (i)  and 
(2). 

Let  d^-\-  d^  =  d  =  the  total  depth  of  the  girder. 

Hence,  from  similar  figures, 


s,      R  +  d,  ^     s,      R-d, 

1  =  -R-     ^''^     1  =  -R- 


Also, 


.  -y,  —  s^  _  d,  -|-  d^       d 
"       I      ~      R      -R' 


/..?,-/      d,  /.    V  -  s,      d^ 

E~     I     ~R     ^""^     E"     I     ~'R' 


E      ~      I      ~      R      ~  R' 


''■>nHI 
'■        i 

'j 

■  i 
■J 

i 

leep  and  6 
deflect  the 


12 


Ex.  5.  A  truss  of  span  120  ft.  and  15  ft.  deep  is.  strained 
so  that  the  flange  tensile  and  compressive  unit  stresses  are 
10,000  and  8000  lbs.,  respectively.  Find  the  deflection,  and 
difference  of  length  between  the  extreme  fibres. 

lOOOO+Sooo  _  .y,  —  .f,  _  15 


30000000 


120 


.*.  .y,  —  J,  =  .864  in.,     and     R  =  25,000  ft. 


Hence  also    D  = 


(120)' 


8  X  25000 


=  .864  in. 


ID.  Camber. — Owing  to  the  play  at  the  joints,  a  bridge- 
truss,  when  first  erected,  will  deflect  to  a  much  greater  extent 


':! 


388 


THEORY  OF  STRUCTURES, 


than  is  indicated  by  theory,  and  the  material  of  the  truss  will 
receive  a  permanent  set,  which,  however,  will  not  prove  detri- 
mental to  the  stability  of  the  structure,  unless  it  is  increased 
by  subsequent  loads. 

If  the  chords  were  made  straight,  they  would  curve  down- 
wards, and,  although  it  does  not  necessarily  follow  that  tlic 
strength  of  the  truss  would  be  sensibly  impaired,  the  appear- 
ance would  not  be  pleasing. 

In  practice  it  is  usual  to  specify  that  the  truss  is  to  have 
such  a  camber,  or  upward  convexity,  that  under  ordinary  loads 
the  grade  line  will  be  true  and  straight. 

The  camber  may  be  given  to  the  truss  by  lengthening  the 
upper  or  shortening  the  lower  chord,  and  the  difference  of 
length  should  be  equally  divided  amongst  all  the  panels. 

The  lengths  of  the  web  members  in  a  cambered  truss  are 
not  the  same  as  if  the  chords  were  horizontal,  and  must  be 
carefully  calculated,  otherwise  the  several  parts  will  not  fit 
accurately  together. 

To  find  an  approximate  value  for  the  camber,  etc.  : 

Let  d  be  the  depth  of  the  truss.    .  , 

Let  J, ,  J,  be  the  lengths  of  the  upper  and  lower  chords, 
respectively. 

Let  /", ,  f  be  the  unit  stresses  in  the  upper  and  lower 
chords,  respectively. 

Let  d^ ,  d,  be  the  distances  of  the  neutral  axis  from  the 
upper  and  lower  chords,  respectively. 

Let  R  be  the  radius  of  curvature  of  the  neutral  axis. 

Let  /  be  the  span  of  the  truss. 


6 

R 


-I 


E 


=  ^J      and      -'-  ^ 


/ 


A 


R 


—  — -. —  =  -p  ,  approximately, 


the  chords  being  assumed  to  be  circular  arcs. 

Hence,  the  excess  in  length,  of  the  upper  over  the  lower 
chord, 


ij 


Si-  S,  =  -gr(/,    f  /,)  =  /^-  =  J/. 


STIFFNESS. 


389 


Let  x^ ,  X,  be  the  cambers  of  the  upper  and  lower  chords, 
respectively.  R  -\-  d^  and  R  —  d^  are  the  radii  of  the  upper 
and  lower  chords,  respectively. 

By  similar  figures,  the  horizontal  distance  between  the  ends 

of  the  upper  chord  =  — - — 'l,  and  the  horizontal  distance  be- 


tween  the  ends  of  the  lower  chord 
Hence, 


A-^/„ 


R 


^ J  —^~n  =  ^i  ■  2{R  +  ^1).  approximately, 


and 


/i  R-d,X  ,„       ^  .         , 

I -^ — /I  =  x^ .  2(/c  —  </,),  approximately. 


'       SRV  ^  R 


and 


=^G- 


Hence,  approximately,  the  camber  =  —  =  —-. 

Note. — The  deflection  of  a  well-designed  and  well-built 
truss  is  often  much  less  than,  and  should  never  exceed,  I  inch 
per  100  ft.  of  span  under  the  maximum  load. 

II.  Stiffness. — If  D  is  the  maximum  deflection  of  a  girder 

W  D 

of  span  /  under  a  load    W^  then  y.-,  or  more  usually  y,  is  a 

measure  of  the  stiffness  of  the  girder. 

In  practice,  the  deflection  of  an  iron  or  a  steel  girder,  under 

the  working  load,  should  lie  between  and  -;. — ,  i.e.,  it  is 

^  1200  600 

limited  to  i  or  2  in.  per  100  ft.  of  span,  and  rarely  exceeds 

,  or  1.2  in.  per  icx)  ft.  of  span. 

1000  ^  ^ 

I 

A  timber  beam  should  not  deflect  more  than  — r-,  or  i  in. 

360' 

per  30  ft.  of  span. 


'^      I 


\'\Vl 


i 


I  ! 


11:1 


I  ;■    vi 


' 


■il 


1 


II 


ir^ 


: 


390 


THEORY  OF  STRUCTURES. 


Let  M^  be  the  bending  moment  at  the  most  deflected 
point.    Then 


Also, 


:^.  =  |/. 


R       EI 


Er 


p  being  a  numerical  coefficient  (in  Art.  9,  Ex.  i,/  =  i;  in  Ex. 
Thus 

gives  the  bending  moment  J/,  to  which  the  girder  of  a  speci- 
fied stiffness  -j  may  be  subjected. 

Again,  if  the  material  is  to  bear  a  certain  specified  unit 
stress  /",  the  maximum  bending  moment  J/,  to  which  the 
girder  may  be  subjected  is  given  by  the  equation 

M,  =  ^I  =  -4/, 

c  qd 

q  being  a  numerical  coefficient  less  than  unity,  depending  upon 
the  form  of  the  section, 

Catcris  paribus,  the  ratio  of  depth  to  span  may  be  fixed 
by  making  the  stiffness  and  strength  of  equal  importance.  -Then 

J/j  =  M^ ;     and  therefore 

^^-^)  =  A 
Pl\ll       ad' 


or 


d 


qd 

iipf 


DISTRIBUTION  OF  SHEARING   STRESS. 


391 


{ 


In  practice  the  proper  stiffness  of  a  g;    ler  is  sometimes 

/ 
secured  by  requiring  the  central  depth  to  lie  between  —  and 

-,  its  value  depending  upon  the  material  of  which  the  girder 

8 

is  composed,  its  sectional  form,  and  the  work  to  be  done. 

Example. — A  cast-iron  beam  of  .'•cctangular  section  and  of 
20  ft.  span  carries  a  uniformly  distributed  load  of  20  tons ;  the 
coefificient  of  working  strength  is  2  tons  per  sq.  in. ;  the  stiff- 
ness is  .001  ;  E  is  8000  tons.     Find  the  dimensions  of  the  beam 
viz..  b  the  breadth  and  d  the  depth. 

20.20  /,         bd^      bd' 

M  =  — ^— .  12  =-/=  2-^  =  —  ; 
8  c  63* 


.-.  bd*  =  1800. 


Also, 


20.  20 


8 


D  ,^     E//D\      S.Sooo. bd'   , 

-  .\2  —  M  =--,[-,] .(.001); 

pl\l I        12  .  20.  12     ^       ' 


.-.  bd' 


pr 

27000. 


Hence, 


27000  .  ,     ,       „  . 

d  =  —^ — ■  =  15  m.     and     (9  =  8  m. 
1800         ^ 


kvi\  ! 


m'A 


12.  Distribution  of  Shearing  Stress. — Let  Figs.  300  and 
301  represent  a  slice  of  a  beam  bounded  by  two  consecutive  sec- 


r^A' 


?c 


ix 


B 


L> 


0' 


B' 


Fig.  300. 

tions  AB,  A'B',  transverse  to  the  horizontal  neutral  axis  00'. 
Let  the  abscissae  of  these  sections  with  respect  to  an  origin 


!    I 


f 


392 


THEORY  OF  STRUCTURES. 


V: 


n 


J ; 


in  tlic  neutral  axis  be  x  and  x  -\-dx,  so  that  the  thickness  of 
the  shce  is  dx. 

In  the  Hmit,  since  dx  is  indefinitely  small,  corresponding 
linear  dimensions  in  the  two  sections  are  the  same. 

Let  /  be  the  moment  of  inertia  of  the  section  AB  (or  A' B' 
in  the  Hmit)  with  respect  to  the  neutral  axis. 

Let  c  be  the  distance  of  A  (or  A'  in  the  limit)  from  the 
neutral  axis. 

Let/, ,  /,  be  the  unit  stresses  at  A  and  A',  respectively. 

Consider  the  portion  ACC A'  of  the  slice,  CC  being 
parallel  to  and  at  a  distance  Y  from  the  neutral  axis.  Since 
it  is  in  equilibrium,  the  algebraic  sum  of  the  horizontal  forces 
acting  upon  it  must  be  nil.     These  forces  are: 

The  total  horizontal  force  upon  ACC, 

"     A'C'C\^x\^ 
"        "  "  shear  along  the  surface  CC . 

The  horizontal  force  upon  an  element  PQ  of  thickness  dy 
and  at  a  distance  y  from  the  neutral  axis 


^f.^^dy, 


z  being  the  width  PQ,.     Thus  the  total  horizontal  force  upon 
ACC 


Ifel 


=  :^[^^y::dy)  =  ^^-^{yzdy)  =~  J]'^dy  =  ^Ay, 


A  being  the  area  of  ACC,  and,;/  the  distance  of  the  centre  of 
gravity  of  this  area  from  00'. 

Similarly,  the  total  horizontal  force  upon  A'C'C 


=  -cj^y'^y = 


fiA- 


-L-.A 


y- 


'/mi 


.  r 


DISTRIBUTION  OF  SHEARING  STRESS. 


393 


kncss  of 

iponding 

(or  A'E 

from  the 

lively. 
C    being 
is.     Since 
\tal  forces 


ickness  dy 


[force  upon 


Hence,  (— '  —  ~\Ay  =  difference  of  the  horizontal  forces 
^'^         ''  upon  A CCiiud  A' C'C, 

=  horizontal  shear  along  CC, 
=  qwdx ; 
<]  being  the   intensity  of   this  shear,  and  tv  the  width  of  the 
section  at  CC. 

Let  il/  and  M  —  r/J/  be  the  bending  moments  at  the  two 
consecutive  sections  AB,  A'B'.     Then 


M=--I    and    M-dM=^--I, 
c  c 


and  therefore 


dM  =  f^'  -  ^)l. 


Hence, 


dM  .-  , 

—  -  Ay  =  qwdx, 


or 


since 


dMAy       S  .- 
^^^  ^  'dx  T  ^  1^'^ 

-—-  =  shearing  force  at  the  section  AB  =  S. 
dx  " 

Example  i.  So/id  rectangular  section  of  zuidth  b. 


-Ay, 

12 

e  centre  of 

or 

. = liy  -  y). 

B 
Fig.  302. 


./ 


/ 


and  the  intensity  of  the  shear  at  any  point  of  AB  may  be  rep- 
resented by  the  horizontal  distance   of   the   point    from   the 

3^" 


parabola  A  VB,  having  its  vertex  at  V,  where  OV  = 


4  be 


ii 


nliU 


a 


!l 


»! 


rr 


r 


i 


1 


394  THEORY  OF  STRUCTURES. 

The  inaximum  intensity  of  shear  is  at  0  and  its  value  is 

_  3    5 

The  value  of  the  average  intensity  is 

5 


Qav    = 


b .  2C 


•  •  Qmax.  •  Yati.  •  •    3  •  ^* 

Ex.  2.    /I  hollow  rectangular  section  ;    B  and  2c  being  the 
external  and  B'  and  2c'  the  internal  xvidth  and  depth. 
At  the  neutriil  axis, 

q{B-B')  =  j  I  ^(.'-o+^^'  ic'  -  n  I . 

Thus,  as  in  Ex.  I,  the  intensity  of  shear  is  again  greatest  al 
the  neutral  plane,  i.e.,  when  Y  =  o. 

Ex.  3.  Solid  circular  section  of  radius  c. 


Ay  =^2^  1V  -/dy  =  K^'  -  Vi, 


w  =  2(c'  ~  V')\     and     /  = 


inc* 


nc' 


and  the  intensity  of  the  shear  at  any  point  of  AB  may  be  rep- 
resented by  the  horizontal  distance  of  the  point  from  the  pa- 

AS 


rabola  AVB,  where  OV  = 


37tc' 


Also,  g„,a^,  =  -^—^     and     ^^,.  =  — •,. 

^TtC  7tC 


.--qa 


'max.  •  fav 


4:    3. 


DISTRIBUTION  OF  SHEARING   STRESS. 


395 


Ex.  4.  A  double-flanged  section,  each  of  the  flanges  con- 
sisting oi  five  8-in.  X  i-in.  plates  riveted  to  a  24-in.  X  i-in.  web 
by  tii'o  3-in.  X  3-in.  X  i-in.  angles. 

To  find  the  intensity  of  shear  at  the  surface  of  contact 
between  the  angles  and  the  flange  : 


^^  =  20  X  I3i  =  265  ;     w  =  6i  in. ;    /  =  8975J, 


neglecting  the  effect  of  the  rivet-holes  in  the  tension  flange. 
Hence 


^f! 


q  =  S 


2120 
466739 


Let  S  =  49  tons.     Then  q  =  .2226  ton  per  square  inch. 

Let  the  rivets  have  a  pitch  of  4  in.,  then 

6i 
the  total  shear  on  each  rivet  =  -^  x  4  X  .2226  =  2.8938  tons. 

Let  the  coefificient  of  shearing  strength  be  4  tons  per  square 
inch,  and  suppose  that  the  surfaces  of  the  angle-irons  and  of 
the  flange  are  close  together ;  then 


c    •     ^        2.8938 
area  of  rivet  =  — ^^-  = 

4 


7234  sq.  in., 


and  its  diameter  =  .96  in. 

If  the  surfaces  arc  not  close  together,  so  that  the  rivet  may 
be  subjected  to  a  bending  action,  then,  by  Ex.  3,  the  average 
intensity  of  shear  in  a  section  =1.4  =  3  tons  per  sq.  in.,  and 

hence 


area 


,   ,     ^      2.8938         ^  ^ 
of  rivet  =  — ^^  =  .9646  sq.  in. ; 

3 


its  diameter  is  i.i  in. 


396 


THEORY   OF  STRUCTURES. 


13.  Beam  acted  upon  by  Forces  Oblique  to  its  Direc 
tion,  but  lying  in  a  Plane  of  Symmetry.— In  discussing  the 
equilibrium  of  such  a  beam  the  forces  may  be   resolved  into 
components  parallel  and  perpendicular  to  the  beam,  and  their 
respective  effects  superposed. 


Fig.  304. 

Let  AB  he  the  beam,  P,,Pj,P^,  ...  the  forces,  and  a, ,  a-, , 
a^  their  respective  inclinations  to  the  neutral  axis. 

Divide  the  beam  into  any  two  segments  by  an  imaginary 
plane  MN  perpendicular  to  the  beam,  and  consider  the  seg- 
ment AjVN. 

It  is  kept  in  equilibrium  by  the  external  forces  on  the  left 
of  MN  and  by  the  elastic  reaction  of  the  segment  BA/N  upon 
the  segment  AMN  at  the  plane  AIN. 

The  resultant  force  along  the  beam  is  the  algebraic  sum  of 
the  components  in  that  direction,  of  /*, ,  P^ ,  P, ,  .  .  .  , 

=  /*,  cos  a,  +  P,  cos  a^-\-  ,  .  .  =  ^{P  cos  a). 

It  may  be  assumed  that  this  force  acts  a'ong  the  neutral 
axis,  and  is  uniformly  distributed  over  tht^  sejtion  A/N. 

Thus,  if  A  is  the  area  of  the  section,  -— — ^  is  the  in- 

A 

tensity  of  stress  due  to  this  force. 

Again,  the  components  of  /*,,/*,,  P, ,  ... ,  perpendicular  to 
the  beam,  are  equivalent  to  a  single  force  and  a  couple  at  MN. 

The  single  force  at  AIN  is  the  Shearing  Force,  is  per- 
pendicular to  the  beam,  and  is  the  algebraic  sum  of  P^  sin  or,, 
/!,  sin  a, ,  .  .  .  , 


=  P,  sin  Of,  -\-  P,  sin  a,  + 


=  2{P  sin  a). 


BEAM  ACTED    UPOX  BY  OBLIQUE   FORCES. 


397 


This    force    develops    a    mean     iixttgcutial    unit    stress    of 

1- — '- '  in  MN.  and  deforms  the  bca;ii,  but  so  sliglitly  as  to 

A 

be  of  little  account. 

The  moment  of  the   couple  is  the  algebraic  sum  of  the 

moments  with  respect  to  MN  oi  P,  sin  «, ,  /-',  sin  or, ,  .  .  .  , 

=  /*,  sin  a^/>^  -\-  /\  sin  nr,/),  -f-  .  .  .  =  2{P/>  sin  a), 

/>,,/>,,...   being  respectively  the  distances  of  the  points  of 
application  oi  P, ,  J\  ,  .  .  .  from  MN. 

Now,  ^{P/>  sin  a)  is  the  resultant  moment  of  all  the  external 
forces  on  the  left  of  MN,  for  the  resultant  moment  of  the  com- 
ponents along  the  beam  is  evidently  ni/.     Hence, 


^{P/>  sin  a)  =  A/=^-^/=  -n. 


E 
R 


and 


/,.  =  j:E{Pp?.\n  a), 


is  the  unit  stress  in  the  material  of  the  beam  at  a  distance  y 
from  the  neutral  axis  due  to  the  bending  action  at  MN  oi  the 
external  forces  on  the  segment  AMN. 

Hence,  also,  the  total  \xmt  stress  in  the  material  in  the  plane 
MN  at  a  distance  y  from  the  neutral  axis  is 


± 


^(jP  cos  a)   ,    ^             ■S'(/^cosry)      y  ^,  ^^^    .       ,         ., 
-^^ ±fy=  ±  -^ ±  -j^{Pp  sm  a)  =/;, 


the  signs  depending  upon  the  kind  of  stress. 

It  will  be  observed  that  this  formula  is  composed  of  tivo 
intensities,  the  one  due  to  a  direct  pull  or  thrust,  the  other  due 
to  a  bending  action.  The  latter  is  proportional  to  the  distance 
of  the  unit  area  under  consideration  from  the  neutral  axis.  It 
is  sometimes  assumed  that  the  same  law  of  variation  of  stress 
holds  true  over  the  real  or  imaginary  joints  of  masonry  and 
brickwork  structures,  e.g.,  in  piers,  chimney-stacks,  walls, 
arches,  etc.  In  such  cases  the  loci  of  the  centres  of  pressure 
correspond  to  the  neutral  axis  of  a  beam,  and   the  maximum 


m 


59« 


THEORY  OF  STRUCTURES, 


III; 


and  minimum  values  of  the  intensity  occur  at  the  edges  of  the 
joint. 

Example  i.  A  horizontal  beam  of  length  /,  depth  d,  and 
sectional  area  A  is  supported  at  the  ends,  and  carries  a  weight 
W  d.X.  its  middle  point.  It  is  also  subjected  to  the  action  of  a 
force  H  acting  in  the  direction  of  its  length. 

First.  Let  the  line  of  action  of  //  coincide  with  the  axis  of 
the  beam. 

The  intensity  of  the  stress  in  the  skin  at  the  centre 


But    coid,    and    /  =  Ak"  a  Ad\ 


-  <xAd=z  — , 
c  n 


n  being  a  coefficient  depending  upon  the  form  of  the  section. 

If  the  section  is  a  circle,  «  =  8 ;    if  a   rectangle,  n  —  6. 
Hence,  i 

,      ,  .                        HI        nW  l\ 
the  skm  stress  =  ±  -^  1 1  H 77  -^ ), 


since  M  = 


Wl 


If  the  load  W  is  uniformly  distributed,  M  = 


m 

8  ■ 


Thus,  a  very  small  load  on  the  beam  may  considerably  in 
crease  the  intensity  of  stress,  and  this  intensity  will  be  still 
further  increased  by  the  deflection  of  the  beam  un^ler  its  load, 
so  that,  in  order  to  prevent  excessive  straining,  it  is  often 
necessary  to  introduce  more  supports  than  are  actually  required 
to  make  the  beam  sufficiently  stiff. 

Second.  If  the  line  of  action  of  //"  is  at  a  distance  h  from 
the  neutral  axis,  an  additional  bending  moment  Hh  will  be  in- 
troduced. 


!ii 


'i  m 


BEAM  ACTED    UPON  BY  OBLIQUE  FORCES. 


399 


Ex.  2,  The  inclined  beam  OA,  carrying  a  uniformly  dis- 
tributed load  of  w  per  unit  of  length,  is  supported  at  A  and  rests 
against  a  smooth  vertical  surface  at  O. 

The  resultant  weight  wl  is  vertical 
and  acts  through  the  centre  C  of  OA  ;  ^|; 
the  reaction  7?,  at  O  is  horizontal. 

Let  the  directions  of  wl  and  /?,  meet 
in  B.  For  equilibrium,  the  reaction  R^ 
at  A  must  also  pass  through  B. 

Let  the  vertical  through  C  meet  the 
horizontal  through  A  in  D. 

The  triangle  ABD  is  a  triangle  of  forces  for  the  three  forces 
which  meet  at  B. 


Fig.  305. 


R^ 
ivl 


AD       AD 


BD 


T^r-—  ~  cot  a, 

2.  DC       2  ' 


a  being  the  angle  OAD.     Hence 


_       wl 
/t,  =  —  cot  a. 
'        2 


Consider  a  section  MN,  perpendicular  to  the  beam,  at  a  dis. 
tancc  X  from  O. 

The  only  forces  on  the  left  of  MN  are  /?,  and  the  weight 
upon  OM.    This  last  is  ivx,  and  its  resultant  acts  at  the  centre 

of  OM,  i.e.,  at  a  distance  -  from  MN. 

2 

The  component  of  R,  along  the  beam 

wl  cos'  a 


=  R,  cos  a  = 


2    sm  a 


The  component  of  Ri  perpendicular  to  the  beam 


r>     .  ^^ 

=  A,  sm  a  =  —  cos  a. 
'  2 


,  jp;  Ji  j 


i  * 


400 


THEORY  OF  STRUCTURES. 


■|.' 


The  component  of  zvx  along  the  beam  =  wx  sin  a. 

The  component  of  w-ir  perpendicular  to  the  beam  =  wxco?,  ix. 

Hence, 

»r,^       tv/ cos'' a   , 

the  total  compression    at  NM  = : -4-  wx  sm  a-  =  t   •, 

^  2    sin  a    ' 


the  shearing  force  at  MN 


tvl 


=  —  cos  a 


wx  COS  a  =  .S\ 


;  \ 


zul  x 

the  bending  moment  at  MN ■=■  x  —  cos  a  —  -  wx  cos  a  =  J/^; 


and 


Jy    —  A  J      ^ 


,  1  \ 


flii 


5;-: 


Ml 


These  expressions  may  be  interpreted  graphically  as  already 
described,  C".,- ,  ^\  being  represented  by  the  ordinates  of  straigiit 
lines,  and  J/j.  ,_/^  by  the  ordinates  of  parabolas. 

fy ,  for  example,  consists  of  two  parts  which  may  be  treated 

independently.  Draw  0£  and  J/' 
perpendicular  to  OA, and  respectively 
equal  or  proportional  to 

wl  cos"  a  w/  cos'  a       xvl    . 

— :  — ; and  — i  —. +  — r  sni  a. 

2 A  sm   a  2 A  sm  a        A 

Join  EF.  The  unit  stress  at  any 
point  of  tlie  beam  due  to  direct  com- 
pression is  represented  by  the  ordinate  (drawn  parallel  to  (V; 
or  AF)  from  that  point  to  EF. 

Upon  ♦^he  line  GG'  drawn  through  the  middle  point  B  per- 
pendicular to  Ox-i,  take  BG  =  BG\  equal  or  proportional  lo 

ywl* 

_--— -  cos  a.  According  as  the  stress  due  to  the  bending  action 
J     o 

at  any  point  of  the  beam  is  compresswe  or  tensile,  it  is  repre- 
sented by  the  ordinate  (drawn  parallel  to  OE  and  AF)  from 
that  point  to  the  parabola  OGA  or  OG'A  ;  G  and  G',  respec- 
tively; being  the  vertices,  and  GG'  a  common  axis. 


m. 


iili 


■i  nt-ii 


SIMILAR   GIRDERS,— PRINCIPAL   PROPERTIES. 


401 


By  superposing  these  results,  the  parabolas  EHF,  EH  F  are 
obtained,  the  ordinates  of  which  are  respectively  proportional 
to  the  values  of  fy  for  the  compressed  and  stretched  parts  of 
the  beam,  i.e.,  for  the  parts  above  and  below  the  neutral 
surface. 

14.  Similar  Girders. — Two  girders  are  said  to  be  similar 
when  the  linear  dimensions  of  the  one  bear  the  same  constant 
proportion  to  the  corresponding  linear  dimensions  of  the  other. 

Thus,  if  fi,  li' ,  rf,  6',  .\,  A'  are  corresponding  breadths,  depths, 
and  lengths  of  two  similar  girders, 

(i       8       \ 

•-J7  =  — ,  =;  -r-,  =  a  constant  =  ^,  suppose. 

pox 

15,  To  Deduce  the  Principal  Properties  of  Similar 
Girders. — {u)  The  weight  of  a  girder  is  proportional  to  the 
product  of  an  area  and  length,  i.e.,  to  the  cube  of  a  linear 
dimension. 

Hence,  the  weights  of  similar  girders  vary  directly  as  the 
cubes  of  their  linear  dimensions.  Hence,  too,  the  unit  stresses 
iiuist  vary  directly  as  their  linear  dimensions. 

\b)  The  Breaking  Weight  of  a  girder  is  calculated  from  a 

ad 
formula  of  the  form  W  —  ^~r,  ^  being  an  area,  rtfa  depth,  and 

/a  leiigtii. 

d 

Now  J  is  constant  for  similar  girders,  so  that  W  is  propor- 
tional to  a,  i.e.,  to  the  square  of  a  linear  dimension. 

Hence, //tf  Breaking  Weights  of  similar  girders  vary  directly 
as  the  squares  of  their  linear  dimensions. 

Example. — A  girder  resting  upon  two  supports  80  ft. 
apart  is  10  ft.  deep  and  weighs  6  tons.  Determine  the  length 
and  depth  of  a  similar  girder  weighing  48  tons. 


/lengthy^  /dq)thy^  48  ^  ^ 


Hence,  the  length  =  160  ft.  and  the  depth  =  20  ft.  Also, 
the  unit  stresses  are  in  the  ratio  of  10  to  20,  and  the  breaking 
weights  in  the  ratio  of  10'  to  20'. 


I 


402 


THEORY  OF  STRUCTURES. 


l;  \ 


16.  To  Discuss  the  Relations  between  the  Correspond- 
ing Sectional  Areas,  Moments  of  Inertia,  Weights,  Bend- 
ing Moments,  etc.,  of  two  Girders  which  have  the  same 
Sectional  Form  and  are  thus  related  : 

The  forces  upon  the  one  being  P^ ,  P^ ,  I\  ,  .  .  .  with  absciss.e 
x^,x^,x^,  .  .  .  those  upon  the  other  are  nP^,  iiP,,  nP^,  .  .  . 
with  abscissiK  /».r, ,  /.v, ,  /.r, . 

The  spans  and  corresponding  lengths  are  in  the  constant 
ratio  /. 

Corresponding  sectional  breadths  are  in  the  constant  ratio ^. 

Corresponding  sectional  depths  are  in  the  constant  ratio  r. 

Let  A,  A'  be  corresponding  sectional  areas  ; 


/,/' 

ti 

«« 

moments  of  inertia  ; 

Q,Q', 

<( 

l( 

weights ; 

S,S' 

<i 

l< 

shearing  forces  ; 

M,M' 

K 

M 

bending  moments; 

fj' 

<i 

<( 

flange  unit  stresses  ; 

s,s' 

« 

« 

web  unit  stresses ; 

R,R' 

II 

It 

radii  of  curvature; 

^,J' 

II 

« 

deflections  ; 

W,  IV' 

II 

<« 

breaking  weights. 

ij.- 


.'.  (a)  A  <x  product  of  a  breadth  and  depth ; 

.-.  A'  =  Aqr. 
{/3)  I  <x  product  of  a  breadth  and  the  cube  of  a  depth ; 

.-.  /'  =Iqr\ 
{y)  Q  oc  product  of  a  length,  breadth,  and  depth  ; 

•  '.Q'=  Qn=  Qpqrp, 

p  being  the  ratio  of  the  specific  weights  of  the  materials  of  the 
girders. 

If  the  materials  are  the  same, 

/o  =  I     and     n  —  pqr. 


SIMILAR  GIRDERS. 
{S)  S'  ^  Sn  =  Spqrp,  for,  from  (;/),  n  =  pqrp. 
(e)  M  is  the  product  of  a  force  and  a  length 
.'.  M'  —  Mnp  =  Mp^qrp, 


(0 


/=f    and    /' 


/'      c' M'  I 


c'M' 


{V) 


/       c  M  I  '^  qr      r 


S  .      ,       S' 

s  =  ^     and     ^  =^\ 


n 


E      f 


s;_A_ 

S  A'~  qr 
E'      f 


■  Pfi. 


403 


(6^)   T)-  =  -      and      -^  =  -- ,  ^  and  £"'   being  the   coef- 


R 


R 


ficients  of  elasticity  of  the  respective  girders ; 


R  ~  f  c  E  ~  E  ^  ffi  ~  E  p'  p' 


(a  length)' 

(0   J  is  proportional  to  — j- r :: —  J 

^  '  r    r  radius  of  curvature 


J' 


R 


Ef 


•'•   A"^  R'  ~  E'  r'^' 


{k)  W\s  proportional  to 


the  product  of  an  area  and  depth 


a  length  ' 

W  _qr .r  _qr^ 

Hence,  the  values  of  A' y  I',  Q',  . .  ,  may  be  derived  from 
tho;e  of  A,  /,  Q,  •  ■  •  hy  means  of  certain  constant  multipliers. 


Hii 


Ir 


■I 


m 


1  *  ! 


t 


1. 

H 


'i 


fi- 
ll 


404 


TUKORY  OF   STRUCTURES. 


Cor.    I.  If  the   two  girders  arc  similar  and   of   the  same 
material, 


p^q 


jA,    E  =  E',    and     /o=  i. 


Hence, 


from  {y),    Q'  —  Qi-t*,   and   the   weights   vary   directly  as  the 

cubes  of  the  linear  dimensions  ; 

"     (e),    M'  =  Mu\  ar.d  the  bending  moments  vary  dircctl) 

a.i    Nc  fourth  powers  of  the  linear  di- 
me., 'ions; 

/  ^' 

"     (C)  and  (?;),  —  -:  /t  •      - ,  ,,  ,J  ihe  flange  unit  stresses  vary 

y  "* 

directly  as  the  web  unit  stresses; 

A' 
"     (i),      -,-  =z  /I*,  and   the  deflections  vary  directly  as  the 

squares  of  the  linear  dimensions ; 

W 
"     (/<■)  -jr/  =  /i° ,  and  the  breaking  weights  vary  directly  as 


W 


the  squares  of  the  linear  dimensions. 


Cor.  2.  Let  the  girders  be  of  the  same  material,  of  equal 
length,  of  equal  rectangular  sectional  areas,  and  equally  loaded. 

Let  b,  b^ ,  and  d,  d^ ,  respectively,  be  the  breadths  and 
depths  of  the  girders.     Then 


Hence, 


^,  =  qb    and     </,  =  rd, 

b^d^  =  qrbd. 
But  b^d^  =  bd\  .  .  qr  =  1.     Also,/  =  i. 


11^. 


.,!|  i 


ALLOWANCE  FOR  THE   WEIGHT  OF  A  BEAM, 


405 


Thus  from  C» 


/ 


R'  A 

'•     Q  and  /,     -^-  =  r'  =  j, ; 


W._     _f 


If  d^  —  ^,     then  /;,  =  d,    and    ^  =  ■/• 


„  r    d       ^   R'    (by 

Hence,  ~^=-t,    'I'ld      77 

f      0  R 


=  g-(-^)' 


17.  To  make  Allowance  for  the  Weight  of  a  Beam. — A 

beam  is  sometimes  of  such  length  that  its  weight  becomes  of 
imi)oitance  as  compared  with  the  load  it  has  to  carry,  and 
must  be  taken  into  account  in  determining  the  dimensions  of 
the  beam. 

The  necessary  provision  may  be  made  by  increasing  the 
cJui/Zi  of  the  beam  designed  to  carry  the  external  load  alone, 
the  width  being  a  dimension  of  the  first  order  in  the  expression 
for  the  clastic  moment. 

Assume  that  tiie  weight  of  the  beam  and  the  external  load 
are  reduced  to  equivalent  uniformly  distributed  loads. 

Let  Wg  be  the  external  load. 


W 

b 
B 


breadth  of  a  beam  designed  to  support  this 

load  only, 
weight  of  the  beam, 
total  load,  the  weight  of   the   beam  being 

taken  into^account. 
corresponding  breadth  of  the  beam, 
weight     "     "        «' 


Then      W-B=  W,, 


and 


B 
B. 


W 
IK 


W, 


W-  B 


W,-B.-  Vv\-B: 


I 


4o6 


THEORY  OF  STRUCTURES. 


%>  _ 


Hence,  b 


Wb  IV  B  W 


W,-B, 


IV.- b: 


IV  -  />; 


Example. — Apply  the  preceding  results  to  a  cast-iron 
girder  of  rectangular  section  resting  upon  two  supports  30  ft. 
apart.  The  girder  is  12  in.  deep  and  carries  a  uniformly  dis- 
tributed load  of  30,000  lbs. 

Take  4  as  a  factor  of  safety  ;  b.  is  given  by 


120000 


C 


b^ 


where  C 
Hence, 


30,000  lbs.,    d=-  \2  in.,     and     /  =  360  in.  ; 
.-.*,=    5  in. 

«       5  X  12  ,      „ 

B,  =  ———  X  30  X  450  =  5625  lbs. ; 
144 

W,  —  B,  =  30000  —  5625  =  24,375  lbs. ; 


m 


30000.5    ^  „ , 

b  = =  6A  in. ; 

24375  ^^ 


^=  ^^  ^  5625  =  6923tV  lbs. ; 

24375 

W=  M^.  +  ^  =  36,923tV  lbs. 


:  ri 


■r, 


Ml 


':'. 

1 

4 

aJ 

EXAMPLES. 


bar  is  bent 


tli( 


of 


ircle  of 


ft.  di 


Iiameter: 

the  coefficient  of  elasticity  is  30,000,000  lbs.  Find  the  moment  uf  resist- 
ance of  a  section  of  the  bar  and  the  maximum  intensity  of  stress  iji  the 
metal,  {a)  when  the  bar  is  round  and  i  in.  in  diameter,  {d)  when  the  bar 
is  square  having  a  side  of  i  inch. 

If  the  metal  is  not  to  be  strained  above  lo.ooo  lbs.  per  sq.  in.,  find  (c) 
the  diameter  of  the  smallest  circle  into  which  the  bar  can  be  bent. 

/Ins. — («)^|4?r  in. -lbs.;  sooolbs.  (/>)  8334- in. -lbs.;  5000  lbs.   (c)  250  ft. 

2.  A  piece  of  timber  10  ft.  long,  12  in.  deep,  8  in.  wide,  and  having  a 
working  strength  of  1000  lbs.  per  sq.  in.,  carries  a  load,  including  its 
own  weight,  of  lu  lbs.  per  lineal  foot.  Find  the  value  of  7f,  (a)  when  the 
timber  acts  as  a  cantilever ;  {i>)  when  it  acts  as  a  beam  supported  at  the 
ends.  Find  (<.)  stress  in  material  3  in.  from  neutral  axis  at  fixed  end  of 
cantilever  and  at  middle  of  beam. 

A/is. — (a)  320  lbs.;  (d)  1280  lbs.;  (c)  500  lbs.  per  sq.  in. 

3.  Is  it  safe  for  a  man  weighing  160  lbs.  to  stand  at  the  ceiitre  of  a 
spruce  plank  10  ft.  long,  2  in.  wide,  and  2  in.  thick,  supported  by  vertical 
ropes  at  the  ends.'  The  safe  working  strength  of  the  timber  is  1200  lbs. 
per  sq.  in. 

Alts.  No  ;  the  maximum  safe  weight  at  the  centre  is  53^  lbs. 

4.  Compare  the  uniformly  distributed  loads  which  can  be  borne  by 
two  beams  of  rectangular  section,  the  several  linear  dimensions  of  the 
one  being  «  times  the  corresponding  dimensions  of  the  other.  Also 
compare  the  moments  of  resistance  of  corresponding  sections. 

Ans.  «' ;  «'. 

5.  A  cast-iron  beam  of  rectangular  section,  12  in.  deep,  6  in.  wide,  and 
16  ft.  long,  carries,  in  addition  to  its  own  weight,  a  single  load  P ;  the 
coefhcient  of  working  strength  is  2000  lbs.  per  sq.  in.  Find  the  value  of 
P  when  it  is  placed  (a)  at  the  middle  point;  {d)  at  2i  ft.  from  one  end. 

Ans.— (a)  847 s\hs.;  {/>)  11,300  lbs. 

6.  A  round  and  a  square  beam  of  equal  length  and  equally  loaded  are 
to  be  of  equal  strength.  Find  the  ratio  of  the  diameter  to  the  side  of  the 
square.  Ans.   |/s6  :  ^33. 

407  , 


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i   . 


II 


408 


THEORY  OF  STRUCTURES. 


7.  Compare  the  relative  strengilis  of  two  beamsof  the  same  length  and 
material  (<i)  when  the  sections  are  similar  and  havi;  areas  in  the  ratio 
of  I  to  4 ;  (lA  when  one  s(;ction  is  a  circle  and  the  other  a  square,  a  side  of 
the  latter  being  eiiual  to  the  diameter  of  the  former. 

Am.  (rt)  I  to  8 ;  (b)  56  to  33. 

8.  Compare  the  strength  of  a  cylindrical  beam  with  the  strength  of 
the  strongest  (a)  rectangular  and  {b)  square  beam  that  can  be  cut  from  it. 

Ans.  {a)  1 12  :  99  1/3";  {b)  33  ^  '4  Va^ 

9.  A  boiler-plate  tube  36  ft.  long,  30  in.  inside  diameter,  weighs  4200 
lbs.  and  rests  upon  supports  33  ft.  apart.  Find  the  maximum  mtensiiy 
of  stress  in  the  metal.  What  additional  weight  may  be  suspended  from 
the  centre,  assuming  that  the  stress  is  nowhere  to  exceed  8000  lbs.  per 
sq.  in.  } 

Ans.  T\\\  lbs.  per  sq.  in.;  18,854 |g J  lbs. 

10.  Compare  the  relative  strengths  of  two  rectangular  beamsof  equal 
length,  the  breadth  {b)  and  depth  (d)  of  one  being  the  depth  (/;)  and 
breadth  ((i)  of  the  other.  Ans.  d'  :  l)\ 

11.  A  yellow-pine  beam  14  in.  wide,  15  in, deep,  and  resting  upon  sup- 
ports 13  in.  apart,  was  just  able  to  bear  a  weight  of  34  tons  at  the  cen- 
tre. What  weight  will  a  beam  of  the  same  material,  of  45  in.  span  and 
S  in.  square,  bear  ?  Ans.  \\\  tons, 

12.  A  cast-iron  rectangular  girder  rests  upon  supports  12  ft.  ai);irt 
and  carries  a  weight  of  2000  lbs.  at  the  centre.  If  the  breadth  is  (w*- 
//(«^the  depth,  find  the  sectional  area  of  the  girder  so  that  the  intensity 
of  stress  may  nowhere  exceed  4000  lbs.  per  sq.  in. 

Ans.   iS  sq.  in.;  if  weight  of  girder  is  to  be  taken  into  account, 
the  depth  d  is  given  by  d"^  —  1.0125c/'  —  216  =;  o. 

13.  Find  the  depth  of  a  wrought-iron  girder  6  in.  wide  which  might  be 
substituted  for  the  cast-iron  girder  in  the  preceding  question,  the  coeffi- 
cient of  strength  for  the  wrought-iron  being  8000  lbs.  per  sq.  in. 

Ans.  4.762  in.;  if  weight  of  girder  is  to  be  included, the  depth  li 
is  given  by  d'  —  .  54c/'  —  108  =  o. 

14.  An  oak  beam  of  circular  section  and  22  ft.  long  is  strained  to  tlie 
elastic  limit  (2  tons  per  sq.  in.)  by  a  uniformly  distributed  load  of  22'! 
tons.  Find  the  diameter  of  the  beam.  What  load  2  ft.  from  one  end 
would  strain  the  material  to  same  limit.'  Ans.  7  in.;  Zi^^^  tons. 

1 5.  A  uniform  beam  of  weight  W^i  crossing  a  given  span  can  bear  a  uni- 
fornily  distributed  load  JV,.  What  load  may  be  placed  upon  the  same 
beam  if  it  crosses  the  span  in  n  equal  lengths  supported  at  the  joints  by 
piers  whose  widths  may  be  disregarded  .-'        Ans.  «'(  ffi-f-  PV-,)  —  JVi. 


m 


EXAMPLES. 


409 


16.  A  flat  spiral  spring  .2  in.  wide  and  .03  in.  thick  is  subjected  to  a 
bctuliii},'  inoiiu  lit  of  10  in. -lbs.  Find  its  radius  of  curvature,  E  bciii^ 
36.000,000  lbs.  Ans,  1.62  in. 


17.  Determine  the  diameter  of  a  solid  round  wrought-:ron  beam  rest- 
ing upon  supports  60  in.  apart  and  about  to  give  way  under  a  load  of  30 
ions  at  14  in.  from  one  end.  Take  5  as  a  factor  of  safety  and  8960  lbs. 
per  sq.  in.  as  the  safe  working  intensity  of  stress. 

Ans.  5.47  in.;  if  weight  of  beam  is  taken  into  account,  the  diam- 
eter {(/)  is  given  by  2019584—  I375(/'—  12320</'  =0. 

18.  A  wrought-iron  bar  li  in.  wide  and  20  ft.  long  is  fixed  at  one  end 
and  carries  a  load  of  500  lbs.  at  the  free  end.  Find  the  depth  of  the  bar. 
so  that  the  stress  may  nowhere  exceed  10,000  lbs.  per  sq.  in. 

Ans.  6.928  In.;  if  weight  of  bar  is  included,  the  depth  d  is  given 
by  ./ »  -  4.8r/  -  48  =  o. 


19.  Compare  the  moments  of  resistance  to  bending  of  a  rectangular 
section  and  of  the  rhomboidal  and  isosceles  sections  which  can  be  in- 
scriixMJ  in  the  rectangle,  the  base  of  the  triangle  being  the  lower  edge 

Ahs.  6  :  I  :  I  or  6  :  I  :  2. 


of  tlie  rectangle. 


30.  A  stress  of  i  lb.  per  sq.  in.  produces  a  strain  of  s^ril^nm  '"  a  beam 
i:  in.  square  and  20  ft.  between  supports.  Find  the  radius  of  curvature 
and  the  central  deflection  under  a  load  of  2000  lbs.  at  the  middle  point. 

A/ts.   2400  ft. ;  J  in. 

21.  A  piece  of  greenheart  139  in.  between  supports,  9  in.  wide  and  8 
to  account,       ^     in.  deep,  was  successively  subjected  to  loads  of  4.  8,  and    16  tons  at  the 

centre,  the  corresponding  deflections  being  .32   in.,  .64  in.  and   1.28  in. 
Find  /fand  the  total  work  done  in  bending  the  beam. 

Wliat  were  the  corresponding  inch-stresses  at  |  of  the  depth  of  the 
beam.>  Ans.  £=  55826821V;  13.44  inch-tons;  H  lb.,  J^  lb.,  \Y  'b. 

22.  The  effective  length  of  the  Conway  tubular  bridge  is  412  ft.  ;  the 
effective  depths  of  a  tube  at  the  centre  and  quarter  spans  are  23.7  ft.  and 
32.25  ^^■'  respectively  ;  the  sectional  areas  of  the  top  and  bottom  flanges 
are  respectively  645  sq.  in.  and  536  sq.  in.  at  the  centre  and  506  sq.  in. 
and  461  sq.  in.  at  the  quarter  spans  ;  the  corresponding  sectional  areas 
of  the  web  are  257  sq.  in.  and  241  sq.  in.  Assume  the  total  load  upon  a 
tube  to  he  equivalent  to  3  tons  per  lineal  foot,  and  that  thecontinuity  of 
tiie  web  compensates  for  the  weakeniiif;  of  the  tension  flange  by  the 
rivet-holes.  Find  the  flange  stresses  and  the  deflection  at  the  centre  and 
quarter  spans,  E  being  24,000,000  lbs.     V/hat  will  be  the  increase  in  the 


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^ 


410 


THEORY  OF  STRUCTURES. 


central  Range  stresses  under  a  uniformly  distributed  live  load  of  J  ton 

per  lineal  foot? 

/ 
Ans.  At  centre  — -  =  181485  ;  /j  =  4-3799  tons  per  sq.  in.  ; 
'+4  /.  =  3.9326 

Deflection  =  8.33  in. 

At  quarter  span  -~  =  132774  ;  ft  =  1.414       "      " 
'44  /,=  1.256        "       " 

Deflection  =  6.25  in. 
The  stresses  and  deflections  are  increased  by  the  live  load  in 
the  ratio  of  5  to  4. 

23.  A  plate  girder  of  64  ft.  span  and  8  ft.  deep  carries  a  dead  load  of  2 
tons  per  lineal  foot.  At  any  section  the  two  flanges  are  of  equal  area,  and 
their  joint  area  is  equal  to  that  of  the  web.  Find  the  sectional  area  at 
the  centre  of  the  girder,  so  that  the  intensity  of  stress  in  tlie  metal  may 
not  exceed  3  tons  per  sq.  in.  The  deflection  of  the  girckr  is  ?  in.  at 
the  centre.     Find  J£  and  the  radius  of  curvature. 

Ans.  128  sq.  in.  ;  15,360  ft.;  25,804,800  lbs. 

24.  Taking  the  coefficient  of  direct  elasticity  at  15,000  tons,  the  coelTi- 

cient  of  latt-ral  elasticity  at  60,000  tons,  and  the  liiiiit  of  elasticity  at  10 

tons,  determine  the  greatest  deviation  from  the  straight  lineof  a  wrought- 

iron  girder  of  breadth  b  and  depth  d.  b"^ 

Ans.  -  ,. 

24000^ 

25.  Find  the  stress  at  the  skin  and  also  at  a  point  4  in  from  the  neu- 
tral axis  in  a  piece  of  10"  x8"  oak,  {a)  with  the  10"  side  vertical;  (b)  with 
the  8"  side  vertical.  The  oak  rests  upon  supports  3  ft.  apart  and  carries  a 
load  of  4900  lbs.  at  its  middle  point.  Also  compare  (r)  the  strength  of  the 
beam  with  its  strength  when  a  diagonal  is  horizontal. 

Ans. — (ii)  330J;  132/jlbs.  per  sq.  in. 

{c)  4  :  4/41  or  5  :  /^li^. 

26.  Find  the  uniformly  distributed  load  which  can  be  borne  by  a 
rolled  T-iron  beam,  6"  X4"  x  i",  10  ft.  long,  fixed  at  one  end  and  free  at 
the  other,  the  coefficient  of  strength  being  10,000  lbs.  per  sq.  in. 

Ans.  438  lbs. 

27.  One  of  the  tubes  of  the  Britannia  bridge  has  an  efTective  length  of 
470  ft.,  depth  of  27^  ft.,  and  deflects  12  in.  at  the  centre  unuer  a  uniformly 
distributed  load  of  1587  tons.  Find  ^and  the  central  flange  stresses,  the 
sectional  areas  of  the  top  flange,  bottom  flange,  and  web  being  648  sq.i".. 
585  sq.  in.,  and  302  sq.  in.,  respectively. 

Ans.  E  =  22,910,496  lbs.;  /(  =  5.37  tons  per.  sq.  in.; 
•     .  .  ■.  /c=4.8i ' 


EXAMPLES. 


411 


:8.  Find  the  moment  of  resistance  to  bending,  of  a  steel  I-beam,  each 
flange  consisting  of  a  pair  of  3-in.  x  3  in.  x  i  in.  angle-irons,  riveted  to 
a  12  in.  X  I  in.  web.  the  coefficient  of  strength  being  5  tons  per  sq.  in. 
.1  load  will  the  beam  <  arry  at  5  ft.  from  one  end,  its  span  being  20 
.t. .''  Find  the  central  deflection,  and  also  the  deflection  at  tlie  loaded 
point,  E  being  I5,cxx)  tons. 

Ans.  287  J  J  in. -tons  ;  6j2J  tons  disregarding  weight  of  beam,  or 
l^\ii  tons  if  weight  of  beam  is  talien  into  account ;  deflection 
at  centre  —  \  in.,  at  loadeJ  point  =  ^^  in. 

29.  A  shaft  5i  in.  deep  x  5  in.  wide  x  98  in.  lonq;  has  one  end  abso- 
luteiv  ti.xed,  while  at  the  other  a  wheel  turns  at  t!ie  rate  of  270  revolutions 
per  minute ;  a  weight  (jf  200  lbs.  is  concentrated  in  the  rim,  its  C.  of  G. 
being  2{  ft.  from  the  axis  of  the  shaft.  Find  the  maximum  stress  in  the 
material  of  the  shaft,  and  also  find  the  maximum  deviation  of  the  shaft 
from  the  straight,  E  bring  27,000,000  lbs. 

30.  The  square  of  the  radius  of  gyration  of  the  equal-flanged  sect  ui 
of  a  wrought-iron  i^'irder  of  depth  d  is  i^//";  the  area  of  the  sei  uon 
—  IcC  ;  tiie  span  =  50  ft.     In  addition  to  its  own  weight  it  carries  a  uiii- 

ily  distributed  load  of  Ij'y  lbs.  per  lineal  foot;  the  maximum  intensity 
ess  =  10,000  lbs.  per  sq.  in.     Find  the  depth.     Also  determine  the 
ji.^/^«j.  A"  being  25,000,000  lbs.  ^Itis.  3*  in.;  -,fj. 

31.  The  central  section  of  a  cast-iron  girder  is  lo^  in.  deep;  its  web 
area  isy^Tv  times  the  area  of  the  topflangc,  and  the  moment  of  resistance 
of  the  section  is  360,000  in. -lbs.;  the  tensile  and  compressive  intensities  of 
stress  are  3000  and  7500  lbs.  per  sq.  in.,  respectively.  Find  the  span  and 
load  so  that  the  girder  may  have  a  stiffness  —  .001,  E  being  17,000,000 
lbs. 

Ans.  rti  =  la-j^T-  sq.  in.;  a-i  —  \\\\  sq.  in.;  a^  +  at  =  97{|f  sq. 
in.;  span  =  136  ft.;  uniformly  distributed  load  =  1764}*  lbs. 

32.  A  double-flanged  cast  'ron  girder  14  in.  deep  and  20  ft.  between  sup- 
ports carries  a  uniformly  distributed  load  of  20  tons.  Find  suitable  di- 
mensions for  the  section,  the  tensile  and  compressive  inch-stresses  being 
2  tons  and  5  tons,  respectively.  Also  f.nd  the  s/iffness  of  the  beam,  E 
being  8000  tons. 

Ans.  Let  thickness  of  web  =  i  in.;  rti  =  22||  sq.  in.;  at  =  4tVj 
sq.  in.;  stiffness  =  .001875. 

33.  The  deflection  of  a  uniformly  loaded  horizontal  beam  supported  at 
the  ends  is  not  to  exceed  i  in.  in  50  feet  of  span,  and  the  stress  in  tlie  ma- 
terial is  not  to  exceed  400  lbs.  per  sq.  in.  Find  the  ratio  of  span  to 
depth.  £■  being  1,200,000  lbs.  per  sq.  in.,  and  the  neutral  axis  being  at 
half  the  depth  of  the  beam.  Ans.  20. 

34.  Two  equal  weights  are  placed  symmetrically  at  the  points  of  tri- 
scction  of  a  beam  of  uniform  section  supported  at  the  ends.   These  weights 


412 


TllF.Oh'V  or  STKUCTirUKS. 


are  then  rnudvcd  aiul  oilier  two  cinial  wri^lits  .ire  placed  at  the  qiiaitcr 
spans,  Kind  llio  ratio  of  the  two  sets  of  wci^litH  s<i  that  the  niaxiimiin 
intensity  of  stress  n\ay  he  the  same  in  each  ease.  Also  show  thai  ilic 
st(lf'nt\ts  of  the  heatn  is  the  saino  in  each  case.  Ahs,  3  to  4. 

35,  A  cast-iron  beam  has  n  cruciform  section  with  equal  rihs  2  in, 
thick  unil  4  in.  lonj".  If  the  intensity  of  longitiulinal  shear  at  the  neiiinil 
axis  is  I  ton  per  sq.  in,,  liiul  tiie  total  slieai  which  the  section  cm 
bear,  and  also  find  the  moment  of  resistance,  the  least  coeincieni  ni 
working  tensile  and  compressive  stress  bein^j  t  ton  per  sf|.  in, 

////.v.   59.31  tons;  34.6  tons, 

3f),   If  a  spiral  sprinjf  is  fastened  to  the  l)arrel  so  that  there  is  hd 
ilianyc  of  diicclion   relatively  to  the  barrel,  sliow  that  the  lendencv  In 
unwind  is  directly  proportional  to  the   amount  of  winding  up,     (('(in 
dition  of  [lerfcct  isoolironism.) 

37.  Show  that  the  tiitu/u/ua  0/  t  iipfiortii  any  maleriid  is  18  times  ilic 
load  which  will  break  a  beam  12  in.  long,  I  in.  deep,  and  1  in.  wide  when 
applied  at  the  centre. 

3.S,  Find  the  limiliii);-  leni>tb  of  a  wroujjjht  iron  cylindrical  beam  4  in, 
in  diameter,  tiic  modulus  of  rupture  being  42,ocx)  lbs.  What  nniforinlv 
distributed  lo;id  will  bre.ik  a  cylindrical  beam  of  the  same  material  :o  li, 
lon^  and  4  in.  in  diameter  ?  Ans.  64.8  ft. ;  8800  lbs. 

39.  ,'\  red-pine  beam  irl  ft.  lotiR  has  to  support  a  weight  of  10,000 Ihs 
at  the  centre.  Tiie  section  is  rectangular  and  tlie  dc|Hh  is  twice  tin- 
breadth.  Kind  the  tiansverse  dimensions,  the  modulus  of  rupture  bcin^ 
S500  lbs.,  and  to  being  a  factor  of  safety.  (Neglect  the  weight  of  tin; 
be.im.)  Ans.  h  =  9.84  in.  ;  d  =■  19,68  in, 

40.  A  round  Oak  cantilever  10  ft.  long  is  just  broken  by  a  load  of  Cifx) 
lbs. suspended  from  the  free  end.  Kind  itsdiameter,  the  moduiusof  rup- 
ture being  10,000  lbs.     (Neglect  the  weight  of  the  Oc.im.) 

Alls.  4.185  in. 

41.  Determine  the  breaking  weight  at  the  centre  of  a  cast-iron  beam 
of  6  ft.  s|»au  and  4  in.  square,  the  coefTicicnt  of  rupture  being  30,txx)li)s. 

Ans.  r.6,66r)||  lbs. 

42.  The  flooring  of  a  corn  warehouse  is  supported  upon  ycllow-piiic 
joists  20  ft.  in  the  clear,  S  in.  wide,  10  in.  deep,  and  spaced  3  ft.  centre 
to  centre.  Kind  tlie  height  to  wliich  corn  weighing  48^  lbs.  per  cii. 
ft.  may  be  heaped  upon  the  floor,  10  being  a  factor  of  safety  and  yMo 
lbs.  the  coeflicient  of  rupture.  Ans.  .68  ft. 

43.  A  yellow-pine  beam  14  in.  wide,  15  in.  deep,  and  resting  upon  siip- 
port'j  126  in.  apart,  broke  down  under  a  uniformly  distributed  load  of 
•60.97  tons.     Find  the  coelficient  of  rupture.  Ans.  2731.456. 


1 1  :  ■ 


.mi:^\ 


=  11 

1 

V  'Ah 

!■ 

■;i»lr 

:i 

K.XAMl'LES. 


413 


.1).  Kitirl  tlir  brcakin^r  wrij^'lit  iit  llir  rnifir  f)f  a  ("iiniidiiin  ;isfi  liriim 
\\  ill  wiilc,  3i  in.  (Ircp.  iitid  of  45  in.  spjin,  tlir  crjciririciit  of  iii|i(,iirc  l)i  'w<^ 
7250.  //"'.  4'Al4!r'ii- 

41;.  A  limber  iH'uin  <'>  in.(ir("|i.  3  in.  vvidi',  </i  in.  bntwrcii  sni>|torl'«,  ;iii'! 
wci^liin)^  50  11)H.  per  en,  ft.,  iiiiiki-  down  imdcf  a  W(!i^lit.  f)f  if>,(x)o  Ih, 
;it  ilic  (M-nlrc.     Find  llic  t:(ndli(i(Mil  of  inpUiic.  Ans.  8911^. 

4(1.  A  wrouKlil-iron  bur  2  in.  wide,  4  in.  deep,  nnd  r44  in.  between  Hiip- 
|i(iiis,  carries  a  unifonniy  diHiribiiled  load  VV  in  ndfblion  to  its  own 
wci^lit-  I'ind  W'.  4  beiii^  u  factor  of  safely  and  5o,o(K)  jbn.  the  roefii- 
( iciil  of  nipiiirc.  Aifi.   5235^  ll)H, 

.(;  I'iii<l  tlic  length  of  a  beam  of  Canadian  ash  6  in.  sfjiiare  wliirh 
wiiiild  lircak  under  its  own  wci^,dit  wlien  Hnpportcd  al  llie  ends.  The 
cocliicifiil  of  rupUiic  —  7c)o(j  lbs.,  an<l  llie  wci^ijlil  of  llie  limber  —  30  lbs. 
pel  (II.  (t.  Alls.   230  ft.    • 

4.S.  The  teeth  of  a, cast-iron  wheel  are  3i>  in.  lon^f,  2I  in.  de(|),  and  7 
in.  wide.  What  is  iIk;  breakinj{  weiylil  of  a  tiMith,  the  cocllii  i<-iit  of 
nipliiii' btMnj.;  5000  ll)s.  ?  Ans.   50,625  lbs. 

41;.  A  wroiiKht-iron  bar  4  in.  dee|),  3  in-  wide,  and  rij^idly  fixed  at  one 
iiui  j^ave  way  at  32  in.  from  the  load  when  loarled  witii   1568  lbs.  at  the 


(ice  (lid.     I''ind  the  coefTu'ient  of  rupture. 


fills.  4iKi^. 


50.  A  east-iron  ijeiim  \2  in.  wide  rests  upon  suppfirts  18  fl.  a|)art,  ;uk1 
carries  a  i2-in.  brieU  wall  which  is  I2i  ft.  in  liei).(ht  anri  w<if4hs  .'12  Ib.s. 
per  (11.  fl.  Takiii)^  63,000  a.s  the  modulus  of  rupture  for  a  uniformly 
(lisli  ihiited  load  and  5  as  a  faci or  of  safety,  find  the  depth  of  the  beam, 
[a]  iiej^lectin(,(  ils  weight;  (h)  lakiii)j;  ils  w<;i^hl  into  account. 

Also  (V;  determine  the  depth  of  a  cedar  i)eam  which  mi^ht  be  .sub- 
stiiiiicd  for  the  cast-iron  L,;am,  taking  11,200  lbs.  a.s  the  modulus  of 
luiiliire  for  the  cedar.  Ans.   {a)  C>  in.  ;  (//)  6i  in. ;  (r)  14.23  in. 

51.  A  cast-iron  girder  27^  in.  deep,  rests  upon  support.s  2G  ft.  apfirt. 
lis  l)(»itom  llaiigi^  has  an  area  of  48  sq.  in.  and  is  3  in.  thick.  Find  the 
breaking  weight  at  the  centre,  the  ultimate  tensile  strength  of  the  iron 
hcini;  i5,(wo  lbs.  per  stj.  in.     (Neglect  the  cflect  of  the  web.) 

All.'!.   253,846 ,«,,  lbs. 

52.  A  bcatn  of  rectangular  section,  of  breadth  /i  and  depth  r/,  is  acu-d 
upon  by  a  couple  in  a  plane  inclined  at  45  to  llu;  axis  of  the  secli"ii. 
C(iiiii)are  the  moment  of  rcsi.stance  to  bending  with  that  about  either 
a.xis.  ^^^^    24/2//^    .  2^/^H 

"  2//'  +f/"'  '26'  + a''' 

53.  A  2-in.  wrought-iron  bar  10  ft.  long  is  held  at  the  ends  and  is 
whirled  about  a  parallel  axis  at  the  rate  t>f  50  revolutions  per  minute. 
If  the  distance  between  the  axis  of  the  bar  and  the  axis  of  rotation  is 
JO  ft.,  find  the  maximum  stress  to  which  the  material  is  subjected. 

Ans.   17148.5  lbs.  per  sq.  in. 


m 


414 


THEORY  OF  STRUCTURES. 


\    \ 


1 


W 


■  i 


54.  A  block  of  ice  3  in.  wide  and  4  in.  deep  has  its  ends  resting  upon 
supports  30  in.  apart  and  carries  a  uniformly  distributed  load  of  4800 
lbs.  An  increase  of  pressure  10  the  extent  of  1 125  lbs.  per  sq.  in.  lowers 
the  freezing  point  i"  F.  Assuming  tliat  the  ordinary  theory  (jf  ilexure 
holds  good,  find  the  temperature  of  the  ice.  Ans.  30'  F. 

55.  Find  the  limiting  length  of  a  cantilever  of  uniform  transverse 
section,/  being  the  coefficient  of  strength,  k  the  ratio  of  length  to  deptii, 
and  w  ihe  specific  weight  of  the  material. 

Ans.  — ~y^t  7/ being  a  coefficient  depending  upon  the  form  of 

the  section. 


56.   If  the  beam  in  the  preceding  question  is  to  be  supported  at  its  two 

I  I  52/« 

wk 


ends,  what  will  its  limiting  length  be  ? 


Ans. 


57.  Find  the  limiting  length  of  a  cedar  cantilever  of  rectangular  sec- 
tion, k  being  40,  w  =  36  lbs.  per  cu.  ft.,  and/=  1800  lbs.  per  sq.  in. 

Ans.  60  ft. 

58.  A  steel  cantilever  2  in.  square  has  an  elastic  strength  of  15  tons 
per  sq.  in.  What  must  its  limiting  length  be  so  that  there  may  be  no 
seif  Ans.  23.4  ft. 

59.  Find  the  limiting  length  of  a  wrought-iron  beam  of  circular  sec- 
tion, k  being  64  and  tlie  elastic  strength  8  tons  per  sq.  in.  What  will  iliis 
length  be  if  a  beam  of  I-section,  having,  equal  flange  areas  and  a  web 
area  equal  to  the  joint  area  of  the  flanges,  is  substituted  for  the  circular 
section  ?  Ans.  84  ft. ;  224  fi. 

60.  A  rectangular  cast-iron  beam  having  its  length,  depth,  and 
breadth  in  the  ratio  of  60  to  4  to  i,  rests  upon  supports  at  the  two  ends, 
Find  the  dimensions  of  the  beam  so  that  the  intensity  of  stress  under  its 
own  weight  may  nowhere  exceed  4500  lbs.  per  sq.  in. 

Ans.  /  =  128  ft. ;  <■/  =  SiV  ft. ;  b=  2-,=^  ft. 

61.  A  beam  supported  at  the  ends  can  just  bear  its  own  weight  //'to 

W 
gether  with  a  single  weight  — -  at  the  centre.  What  load  may  be  placed 

at  the  centre  of  a  beam  whose  transverse  section  is  similar  but  vr  as 
great,  its  length  being  n  limesas  great  ?  If  the  beam  could  support  only 
its  own  weight,  what  would  be  the  relation  between  m  and  « } 


I  vi^       in'\  a 

Ans.    IV ;    m  — 

\n         2  j  2 


62.  The  flanges  of  a  rolled  joist  are  each  4  in.  wide  by  \  in.  thick; 
the  web  is  8  in.  deep  by  i  inch  thick.  Find  the  position  of  tlie  neutral 
axis,  the  maximum  intensities  of  stress  per  square  inch  being  kjooo 
lbs.  in  tension  and  8000  lbs.  in  compression.         Ans.  h\  =  3J  ;  //g  =  4I. 


ii^. 


ff^^ 


fiiiil 


EXAMPLES. 


415 


Ling  upon 
i  of  4800 
in.  lowers 
of  liexure 

.  30'  ^''• 
transverse 
1 10  depth, 

le  form  of 


;d  at  its  two 

wk 
ingular  sec- 
sq.  in. 
Ans.  60  ft. 

h  of  I  5  tons 
I  may  be  no 
u.  23.4  ft- 
circular  st'C- 
/hai  will  this 
s  and  a  web 

the  circular 
t. ;  224  ft. 

depth,  and 
Ihe  two  ends, 
ess  under  its 

\b  =  2f,T  ft. 

veight  fTto 

|ay  be  placeri 

ir  but  in-  ai 
Isupportonly 


//i  =  ^  ■ 

\  in.  thick; 
tlie  neii trill 
|being  10000 
;  hi  =  4ft' 


63.  A  continuous  lattice-girder  is  supported  at  four  points,  each  of  the 
side  spans  being  140  ft.  11  in.  in  length,  22  ft.  3  in.  in  depth,  and  weigh- 
ing .68  ton  per  lineal  foot.  On  one  occasion  an  excessive  load  lifted  the 
end  of  one  of  the  side  spans  off  the  abutment.  Find  the  consequent  in- 
tensity of  stress  in  the  bottom  flange  at  the  pier,  wliere  its  sectional 
area  is  127  sq.  in,  Ans,  2.3893  tons  per  sq.  in. 

64.  A  railway  girder  is  101.2  ft.  long,  22.25  ft.  deep,  and  weighs  3764 
lbs,  per  lineal  foot.  Find  the  maximum  shearing  force  and  flange  stresses 
at  25  ft.  from  one  end  when  a  live  load  of  2500  lbs.  per  lineal  foot  crosses 
the  girder. 

65.  A  floor  with  superimposed  load  weighs  160  lbs.  per  sq.  ft.  and  is 
carried  by  tubular  girders  17  ft.  c.  to  c.  and  42  ft.  between  be;  '  ings. 
Find  the  depth  of  the  girders  (neglecting  effect  of  web),  the  safe  inch- 
stress  in  the  metal  being  9000  lbs.,  and  the  sectional  area  of  the  tension 
flange  at  the  centre  32  sq.  in.  Ans.  24.99  '"• 

66.  Design  a  timber  cantilever  of  fl//r<7,rmrt/<?/)/ uniform  strength  from 
the  following  data:  length  =  12  ft.;  square  section;  load  at  free  end 
=  I  ton ;  coefficient  of  working  strength  =  \  ton  per  sq.  in.  What 
must  be  the  dimensions  at  the  fixed  and  free  ends  so  that  tlie  cantilever 
miglit  carry  an  additional  uniformly  distributed  load  of  2  tons  ? 

Ans.  Side  =  15.1    in.  at    fixed    end  and  =  10   in.  at  free  end; 
side  =  19.1    in.  at  fixed  end  and  =  K'9'    •"•)   at   tree   end. 

67.  Show  that  the  curved  profile  of  a  cantilever  of  uniform  strength 
designed  to  carry  a  load  W  at  the  free  end,  is  theoretically  a  cubical 
parabola.  Also  show  that  by  taking  the  tangents  to  the  profile  at  the 
fixed  end  as  the  boundaries  of  the  cantilever,  a  cantilever  of  approxi- 
mately uniform  strength  is  obtained  having  a  depth  at  the  free  end 
oqual  to  two-thirds  of  the  depth  at  the  fixed  end, 

68.  Design  a  wheel-spoke  33  in.  in  length  to  be  of  approximately  uni- 
form strength,  the  intensity  of  stress  being  4000  lbs.  per  sq.  in.  ;  the  load 
u  the  end  of  the  spoke  is  a  force  of  1000  lbs.  applied  tangentially  to  the 
wheel's  periphery,  and  the  section  of  tlie  spoke  is  to  be  {a}  circular,  (p) 
elliptical,  the  ratio  of  the  depth  to  the  breadth  being  i\. 

Ans. — («)     Depth  at  hub  =  6.982    in.,  at  periphery  =  4.634  in. 

(p)  =  9435     =  9'3S  in. 

Breadth  "     "     =  3.774     =  3.74  in. 

69.  A  beam  of  17  ft,  span  is  loaded  with  7,  7,  ir,  and  11  tons  at 
points  1,6,  II,  and  15  ft.  from  one  end.  Determine  the  depths  at  the.se 
points,  the  beam  being  oi  uniform  breadth  and  of  approximately  uni- 
form strength ;  the  coeflicient  of  working  strength  =  2  tons  per  sq.  in.; 
the  depth  of  the  section  of  maximum  resistance  to  bending  =  16  in. 


16065  277  X   i6* 

Ans,   b  =    ._o„  ;  (i\  — 


1088 
and  d\ 


1785 
670  X  16' 
1785     * 


.,       1087  X  16'     .. 


'•ill 


t  i  ' 


Kt 


416 


THEORY  OF  STRUCTURES. 


70.  Design  a  cantilever  10  ft.  long,  of  approximately  \x'cn[or\n?X.xzx\^^\.\\, 
to  carry  a  load  of  4000  lbs.  at  the  free  end,  the  coefficient  of  strenjrih 
being  2000  lbs.  per  sq.  in.,  and  the  section  {a)  a  rectangle  of  constant 
breadth  and  I3  in.  deep  at  the  fixed  end  ;  {b)  a  square. 

How  will  the-  results  be  modified  if  it  is  to  carry  an  additional  uni- 
formly distributed  load  of  4800  lbs.  } 

Ans. — First,   (a)  6  =  10  in.,    li  at    free    end  =  6   in.;    (d)   side 
=  ^  1440  at  fixed  end  and  =    f-'iSo  at  free  end. 
Second,  (a)  h  =  16  in.,  ^/at  free  end  =  6  in. ;  (d)  side  =  -^2304 
at  fixed  end  and  =   1^288  at  free  end. 

71.  Design  a  cantilever  10  ft.  long,  of  constant  breadth,  and  of  /rp- 
proximately  uniform  strength  to  carry  a  uniformly  distributed  load  of  5000 
lbs.  on  the  half  of  the  length  next  the  free  end,  the  intensity  of  stress 
being  2000  lbs.  per  sq.  in.,  and  the  section  a  rectangle  12  in.  deep  at  the 
fixed  end.  VVluit  must  the  dimensions  be  if  1000  lbs.  are  concentrater] 
at  30  in.  from  fixed  end  .' 

Ans.  h  —  (^  in. ;  d  S.X.  centre  =  6.928  in.  ;  at  free  end  =  o. 

b—\Q)  in.  ;  depth  =  8.66  in.  at  7i  ft.  from  free  end,  =6.928 
in.  at  centre,  and  =  o  at  free  end. 

72.  A  gallery  30  ft.  'ong  and  10  ft.  wide  is  supported  by  four  9  in.  by  5 
in.  cantilevers  spaced  so  as  to  bear  equal  portions  of  the  superincumbent 
weight.  What  load  per  square  foot  will  the  gallery  bear,  the  coefficient  of 
working  strength  being  700  lbs.  per  sq.  in. .'  Find  the  deptli  of  cast-iron 
cantilevers  3  in.  wide  which  may  be  substituted  for  the  above,  the  coef- 
ficient of  working  strength  being  2000  lbs.  per  sq.  in.  How  should  the 
depth  vary  if  the  cantilevers  are  to  be  of  uniform  strength  ? 

Ans.  42  lbs.;  d'^  =  18.9;  variation  of  depth  for  cast-iron  canti- 
lever is  given  by  1000^/'-  =  189.1',  x  being  distance  from  free 
end. 

73.  A  span  of  60  ft.  is  crossed  by  abeam  hinged  at  the  points  of  trisec- 
tion  and  fixed  at  the  ends  :  the  beam  has  a  constant  breadth  of  3  in.  and 
is  to  be  of  uniform  strength  ;  the  intensity  of  stress  is  3  tons  per  sq.  in. 
Determine  the  dimensions  of  the  beam  when  a  load  of  j'j  ton  per  lineal 
foot  covers  (a)  the  whole  span  ;  (b)  the  centre  span. 

Ans.— {a)  Depth  at  support  =  20  in.,  at  centre  =  4/20. 
{b)       •'  "       =  i/So  in.,  "        "      =  t^2o' 

74.  In  the  following  examples  determine  the  position  of  the  neutral 
axis,  the  moment  of  resistance  to  bending,  the  resistance  to  shear,  and 
the  ratio  of  the  maximum  to  the  average  intensity  of  shear,  the  co- 
efficients of  strength  being 4^  tons  per  sq.  m.  for  tension  and  compression 
and  34  tons  per  sq.  in.  for  shear. 

(I)  A  rectangle  ^  .r..  wide  and  6  in.  deep. 

Ans.  At  centre  ;   54  m. -tons;  28  tons  ;  3  to  2. 


iMfl 


<  5,(1 


EXAMTLES. 


417 


(II)  A  circular  section  4  in.  in  diameter. 

Alls.  At  centre  ;   28.2  in. -tons;  33  tons;  4  to  3. 

(III)  A  regular  hexagonal  section  with  a  diameter  {a)  vertical,  (<5) 
horizontal,  a  being  a  side  of  the  hexagon. 


Ans. — (a)  At  centre; 


4!"VJ;  ^«v 


3 ;  7  to  5. 


(<J)  At  centre  ;  i5^,».    -^^3^^^.  a""  \  i-z^'i. 
lO  1888 

(IV)  A  triangular  section  6  in.  deep,  with  a  base  6  in.  wide,  the  sides 
being  equal.  Ans.  4  in.  from  vertex  ;  40.5  in.-t(jns  ;  42I  tons  ;  3  to  2. 

(V)  A  double-tee  section  composed  of  a  30-in.  x  f-in.  web  and  four 
anglt-irons  each  5  in.  x  3^  in.  x  f  in. 

Ans.  At  centre  ;  1501.06  in. -tons  ;  22.36  tons;  4.916  to  I. 

(VI)  A  section  having  a  semicircular  top  flange  of  8  in  external 
diameter  and  i  in.  thick,  a  web  14  in.  deep  and  i  in.  thick,  and  a  bottom 
flange  8  in.  wide  and  i  inch  thick. 

(VII)  A  section  having  a  semi-elliptic  top  flange  2  in.  thick,  the  in- 
ternal major  and  minor  axes  being  8  in.  (Jiorizontally)  and  4  in.  {ver- 
tically), respectively,  a  bottom  flange  8  in.  wide  and  2  in.  thick,  and  a 
web  10  in.  deep  and  2  in.  thick. 

(VIII)  A  section  having  a  semi-elliptic  top  flange  2  in.  thick,  the 
external  major  and  minor  axes  being  10  in.  {horizontally)  and  6  in. 
{vertically),  respectively,  a  trapezoidal  web  8  in.  deep  having  a  width  of  3 
in.  at  the  top  and  6  in.  at  the  bottom,  and  a  bottom  semicircular  flange 
of  10  in.  external  diameter  and  2  in.  thick. 

(IX)  The  sections  shown  by  Figs.  307,  308,  and  309. 


v-3'>J 8- »l-3^ 


^ 10 


Fig.  307. 


V io^ * 


Fig.  3og. 


Also  find  the  diameters  of  the  rivets  R  in  Fig.  i,  neglecting  the  weak- 
ening effect  of  the  rivet-holes  in  the  bottom  flange.  What  is  the  ratio 
of  the  maximum  tensile  and  compressive  stresses  in  each  section  } 


m 


4i5 


THEORY  OF  STRUCTURES. 


\.i,'\ 


(X)  A   trapezoidal  section,  the  top  side,  bottom  side,  and  depth  // 
(inches)  being  in  the  ratio  of  i  to  2  to  4. 

Ans.  ^h  from  top  side  ;  7^5^'  in. -tons. 

(XI)  A  section  in  the  form  of  a  rhombus  of  depth  2c  and  with  a  hori- 
zontal diagonal  of  length  2b.  Ans.  \hc'^ ;  ^^bc  ;  9  to  8. 

(XII)  An  angle-iron  2  in.  x  2  in.  x  ^  in. 

.Ins.  Neutral  axis  divides  depth  into  segments  of  jf  in.  and  \\ 
in. ;  WW  in.-tons;  ^§^^  ton;  1334  to  1369. 

(XIII)  A  hollow  circular  section  of  external   radius  C  and  interna! 
radius  C. 


Ans. 


.99 
112 


C'     33 


C'-C 


4  C»  +  CC  +  €•■' 
3 


C        •   4   C'»  +  CC  +  C"  •  3         C'  +  C" 
(XIV)  A  cruciform  section  made  up  of  aflat  steel  bar  10  in.  by  \  in. 
and  four  steel  angles,  each  4  in.  by  4  in.  by  i  in.,  all  riveted  together, 
(Neglect  weakening  eflfect  of  rivet-holes.) 

75.  A  girder  of  21  ft.  span  has  a  section  composed  of  two  equal 
flanges  each  consisting  of  two  3i-in.  x  5-in.  x  \-\n.  angles  riveted  to  a 
39-in.  X  |-in.  web ;  the  cover-plates  on  the  flanges  are  each  12  in.  x  |  in., 
and  the  rivets  in  the  covers  alternate  with  those  connecting  the  angles 
and  web ;  the  pitch  of  the  rivets  is  3^  in.  Find  the  diameter  and  also  firui 
the  maximum  flange  stresses,  {a)  disregarding  the  we  '.!;ening  eflfect  of  thu 
rivet-holes  in  the  tension  flange  ;  (b)  taking  this  eflfect  into  account 

The  load  upon  the  girder  is  a  uniformly  distributed  load  of 
20,800  lbs.  (including  weight  of  girder)  and  a  load  of  50,000  lbs.  concen 
trated  at  each  of  the  points  distant  4^  ft.  from  the  middle  pomt  of  the 
girder. 

Ans.  Diam.  of  rivets  =  .48  in.  if  tight,  =  .54  in.  if  subject  to  flexure. 
{a)  /i  —f-i  =  7762  lbs.  per  sq.  in. 
(b)  /,  =  8248  lbs.  per  sq.  in., /a  =  7847  lbs.  per  sq.  in. 

76.  A  beam  of  triangular  section  12  in.  deep  and  with  its  base  hori- 
zontal can  bear  a  total  shear  of  100  tons.  If  the  safe  maximum  intensity 
of  shear  is  4  tons  per  sq.  in.,  find  tlie  width  of  the  base.       Ans.   6J  in, 

77.  Assuming  that  the  web  and  flanges  of  a  rolled  beam  are  rectangular 

in  section,  determine  the  ratio  of  the  maximum  to  the  average  intensity 

n 
of  shear  in  a  section  from  the  following  data  :   the  total  tiep/A  is  -  times 

the  breadth  of  each  hange,  «  times  the  thickness  of  each  flange,  and  m 
times  the  thickness  of  the  web.  Show  also  that  this  ratio  is  ^  or  V 
according  as  the  area  of  the  web  is  equal  to  the  joint  area  of  the  two 
flanges  or  is  equal  to  the  area  of  each  flange.  How  much  of  the  shear 
in^-  force  is  borne  by  the  web  ?     How  much  by  the  flange  ? 

3(«'  -)-  I2«  —  I2)(«  +  6) 


Ans.  ratio 


2(«'  -f-  i8«'  —  36«  4-  34) ' 


7o)« ;  85^. 


:t  to  flexure. 


EXAMPLES. 


419 


78.  In  a  rolled  beam  with  equal  flanges,  the  area  of  the  web  is  propor- 
tional to  the  «ih  power  of  the  depth.  Find  the  most  economical  distribu- 
tion of  metal  between  the  flanges  and  web,  and  the  moment  of  resistance 
to  bending  of  the  section  thus  designed.  Also  find  the  ratio  of  the  aver- 
age to  the  maximum  intensity  of  shear. 

Ans.  Area  of  each  flange  :  web  area  ::  2«  —  i  :  6  ; 

n 

2  n—J^y' 


B.  M.  =  - 


>        /  being  the  coefficient  of  strength,  5  the  total  area  of 

section,  and  y  the  depth. 
Max.  intensity  of  shear  :  av.  intensity  ::  («  -f  \)(\ii  -(-  i)  :  6«. 

79.  Find  the  moment  of  resistance  to  bending,  the  resistance  to  shear, 
and  the  ratio  of  maximum  to  the  average  intensity  of  a  shear  in  the 
case  of  a  section  consisting  of  two  equal  flanges,  each  composed  of  a 
pairof  5-in.  X  3i-in.  x  f-in.  angle-irons  riveted  to  a  3ii-ia  x  f-in.  web, 
tiie  5-in.  sides  of  the  angles  being  horizontal,  and  4^  tons  per  sq.  in. 
being  the  coefficient  of  strength. 

Ans.   1501.06  in. -tons  ;  22.36  tons  ;  4.916. 

80.  The  floor-beam  for  a  single-track  bridge  is  15  ft.  between  bearings, 
and  each  of  its  flanges  is  composed  of  a  pairof  2|-in.  x  2^-in,  x  f-in. 
angle-irons  riveted  to  a  30-in.  x  f-in.  web.  The  uniformly  distributed 
load  (including  weight  of  beam)  upon  the  beam  is  4200  lbs.,  and  a  weight 
of  1600  lbs.  is  concentrated  at  each  of  the  rail-crossings,  2\  ft.  from 
the  centre.  Find  ia)  the  maximum  flange  stress,  (I))  the  ratio  of  the 
maxiinitin  and  average  intensities  of  shear ;  (f)  the  stiffness,  E  being 
27,000,000  lbs. 

Ans.  (rt)  6523.4  lbs  ;  (b)  2.037;  (c)  .00033. 
24512.59 


/  = 


1024 


neglecting  effect  of  rivet-holes. 


81.  A  beam  36  ft.  betv\'een  bearings  is  a  hollow  tube  of  rectangular  sec- 
tion and  consists  of  a  24ii..  x  ^-in.  lop  plate,  a  24-in.  x  ,}-in.  bottom 
plate,  and  two  side  plates  each  35  in.  x  ^  in.  The  plates  are  riveted 
together  at  the  angles  of  the  interior  rectangle  by  means  of  four  6-in. 
X  4-in.  x^-in.  angle-irons,  the  6-in.  side  being  horizontal.     Determine — 

{a)  The  intensity  of  shear  at  the  surface  between  the  angle-irons  and 
the  upper  and  lower'plates. 

{b)  The  diameter  of  the  rivets,  the  pitch  being  4  in.  and  assuming  an 
effective  width  of  5^  in.  in  shear  per  rivet. 

(<r)  The  total  shearing  strength  of  the  section,  the  safe  intensity  of 
shear  being  3^  tons  per  sq.  in. 

(d)  The  moment  of  resistance  of  the  section,  the  coeflBcient  of  strength 
being  4J  tons  per  sq.  m. 

(e\  The  uniformly  distributed  load  which  the  beam  will  safely  carry. 


■.V 


[iiiiji 


i  I 


t 


420 


THEORY  OF  STRUCTURES. 


Ans. — {a)  .11878  tons  per  sq.  in. 

(b)  .97  in.  if  rivets  are  tiglit,  1. 12  in.  if  liable  to  flexure. 

{c)  i09^Y/ff   tons  disregarding  effect  of  riveting,  los/j^j?, 

tons  having  regard  to  riveting. 
(</)  4o85iin.-tons  disregarding  effect  of  riveting,  3838.9157 

tons  having  regard  to  riveting. 
{e)  75f|  tons  disregarding  effect  of  riveting,  71.09  tons 

having  regard  to  riveting. 

82.  A  cast-iron  channel-beam  having  a  web  12  in.  wide  and  two  sides 
7  in.  deep,  the  metal  being  everywhere  i  in.  thick,  crosses  a  span  of  14 
ft.  If  the  tensile  intensity  of  stress  is  i  ton  per  sq.  in.,  what  uniformly 
distributed  load  will  the  beam  carry  {a)  with  the  web  at  the  bottom  ;  (/;) 
with  the  web  at  the  top.-"  Find  {c)  the  maximum  compressive  intensity 
of  stress  to  which  the  metal  is  subjected,  and  (</)  compare  the  maximum 
and  average  intensities  of  shear.  Also,  [e)  what  should  be  the  area  of 
a  rectangular  section  to  bear  the  same  total  shear  ? 

Ans.  I  —\\o\\  {a)  \\\  tons  ;  (b)  %\\  tons. 

83.  A  beam  of  rectangular  section  and  of  a  length  equal  to  20  times  the 
depth  is  supported  at  the  ends  in  a  horizontal  position,  and  is  subjected 
to  a  thrust  //whose  line  of  action  coincides  with  the  axis  of  the  beam. 
Show  that  the  maximum  intensity  of  stress  at  the  middle  point  will  be 
doubled  by  concentrating  at  that  point  a  weight  ^F" equal  to  one-thirtieth 
oiH. 

84.  The  line  of  action  of  the  thrust  in  a  compression  member  is  at  a 

distance  from  the  axis  equal  to  -th  of  the  least  transverse  dimension. 

r 

Show  that  the  maximum  intensity  of  stress  is  doubled   if  the  section  is 
rectangular  and  r  =  6,  or  if  the  section  is  circular  and  r  =  8. 

85.  A  straight  wrought-iron  bar  is  capable  of  sustaining  as  a  strut  a 
weight  wi.and  as  a  beam  a  weight  Wa  at  the  middle  point,  the  deflection 
being  small  as  compared  with  the  transverse  dimensions.  If  the  bar  has 
simultaneously  to  sustain  a  weight  w  as  a  strut  and  a  weight  w'  as  a 
beam,  the  weight  being  placed  at  the  middle  of  the  span,  show  that  the 
beam  will  not  break  if 


7£'i        , 
W  ■\-  —  W 


<  w,. 


86.  A  metal  beam  is  subjected  to  the  action  of  a  bending  moment 
steadily  applied  beyontf  the  elastic  limit.  Assuming  that  ?.he  metal  acts 
as  if  it  were  perfectly  plastic,  i.e.,  so  that  the  stress  throughout  a  trans- 
verse section  is  uniforvt,  compare  the  moment  of  resistance  to  bendmg 
of  a  section  of  the  beam  with  the  moment  on  the  assumption  that  the 
metal  continued  to  fulfil  the  ordinary  laws  of  elasticity,  {a)  the  section 
being  a  rectangle  ;  {b)  the  section  being  a  circle. 


EXAMPLES. 


421 


87.  A  lattice-girder  of  looft.  span  carries  80  tons  uniformly  distributed; 
the  girder  is  10  ft.  deep  and  the  safe  working  stress  is  4  tons  per  sq.  in. 
If  the  width  of  the  flange  must  be  20  in.  to  carry  the  load  exclusive  of 
the  weight  (jf  the  girder,  what  must  be  the  width  of  the  flange  when  the 
weight  of  the  girder  is  taken  into  account  ? 

88.  A  plate-girder  of  double-tee  section  and  of  80  ft.  span  is  8  ft.  deep 
and  carries  a  uniformly  distributed  load  of  80  tons.  If  the  width  of  the 
lliinge  must  be  12  in.  to  carry  the  load  exclusive  of  the  weight  of  the  gir- 
der, what  must  the  width  be  when  this  weight  is  taken  into  account  } 

89.  If  the  plane  of  bending  does  not  coincide  with  the  plane  of  sym- 
metry of  a  beam,  show  that  the  neutral  axis  is  parallel  to  a  line  joining  the 
centres  of  two  circles  into  which  the  beam  would  be  bent  by  two  com- 
ponent couples  whose  axes  are  the  principal  axes  of  inertia  of  the  section, 
each  couple  being  supposed  to  act  alone. 

90.  The  flanges  of  a  girder  are  of  equal  sectional  area,  and  their  joint 
area  is  equal  to  that  of  the  web.  What  must  be  the  sectional  area  10  resist 
a  bending  moment  of  300  in. -tons,  the  effective  depth  being  10  in.  and 
the  limiting  inch-stress  4  tons.'  •  Ans.  22|  sq.  in. 

91.  The  effective  length  and  depth  of  a  cast-iron  girder  which  failed 
under  a  load  of  18  tons  at  the  centre  were  57  in.  and  5J  in.,  respectively ; 
the  top  flange  was  2.33  in.  by  .31  in.,  the  bottom  flange  6.67  in.  by  .66  in., 
and  the  web  was  .266  in.  thick.  Assuming  that  the  ordinary  theory  of 
flexure  held  good,  what  were  the  maximum  intensities  of  stress  in  the 
flanges  at  the  point  of  rupture  .' 

Ans.  ft  =  12.36  tons  per  sq.  in. ;  /a  =  44.9  tons  per  sq.  in. 

92.  A  railway  bridge  is  supported  upon  two  main  girders  each  of  span 
51  ft.  4  in.  ;  at  the  centre  the  depth  is  6  ft.  6  in.,  the  gross  sectional  area 
of  the  top  flange  27  sq.  in.,  and  of  the  bottom  flange  28  sq.  in.  Assum- 
ing the  efficiency  of  the  tension  flange  is  reduced  onc-fifik  by  the  rivet- 
holes,  find  the  maximum  flange  intensities  of  stress  under  a  uniformly 
distributed  load  of  43  tons.  Also  find  the  uniformly  distributed  rolling 
load  which  will  increase  these  intensities  by  two  tons. 

Ans.  .786  ton  per  sq.  in.  in  compression  ;  .9475  ton  per  sq.  in.  in 
tension;  55^  tons  to  increase  compression;  sg,**/'/?'?  ^o"s  to 
increase  tension. 

93.  A  lattice-girder  of  80  ft.  span  and  8  ft.  deep  is  designed  to  carry  a 
dead  bad  of  5o  tons  and  a  live  load  of  120  tons  uniformly  distributed  ; 
at  the  centre  the  net  sectional  area  of  the  bottom  flange  is  45  sq.  in., 
and  the  gross  sectional  area  of  the  top  flange  56^  sq.  in.  Find  the  po- 
sition of  the  neutral  axis  and  the  maximum  flange  intensities  of  stress. 
If  the  live  load  travels  at  60  miles  an  hour,  what  will  be  the  increased 
pressure  due  to  centrifugal  force? 


nrr 


422 


THEORY  OF  STRUCTURES. 


Ans.  3.546  ft.,  from  top;    1120  lbs.  per  sq.  in.;  8920.35  lbs.  per 

594000 
*  sq.  in. ;  ~~ir~  lbs. 

94.  Determine  the  thickness  of  the  metal  in  a  cast-iron  beam  of  12  ft. 
span  and  8  in.  deep  which  has  to  carry  a  uniformly  distributed  load  oi 
j.ooo  lbs.,  the  section  being  («)  a  hollow  square  ;  (^)  a  circular  annulus. 
The  coefficient  of  working  strength  =  3000  lbs.  per  sq.  in.  Also  lind  tlu- 
limiting  safe  span  of  the  beam  under  its  own  weight. 

Ans.  Neglecting  weight  of  beam,  (a)  .281  in.;  (^)  477  in.  Takiiii; 
weight  of  beam  into  account,  («)  .307  in.  ;  (p)  .534  in.  Liniitin- 
span  =  41.3  ft.  in  {a)  and  =  35.7  ft.  in  {b). 

95.  Determine  suitable  dimensions  for  a  cast-iron  beam  20  in.  decfi. 
at  a  section  subjected  to  a  bending  moment  of  1200  in.-tons  ;  the  coeffi- 
cients of  strength  per  square  inch  being  2  tons  for  tension  and  8  tons  for 
compression.     Take  thickness  of  web  =  /u  in. 

Ans.  Sectional  area  of  tension  flange  =  36  sq.  in.;  of  compression 
flange  =  2J  sq.  in. 

96.  The  thickness  of  the  web  of  an  equal-flanged  I-beam  is  a  certain 
fraction  of  the  depth.  Show  that  the  greatest  economy  of  material  is 
realized  when  the  area  of  the  web  is  equal  to  the  joint  area  of  the 
flanges,  and  that  the  moment  of  resistance  to  bending  is  \fAd,f  being 
the  coefficient  of  strength,  A  the  total  sectional  area,  and  d  the  depth. 

97.  In  a  double-flanged  cast-iron  beam  the  thickness  of  the  web  is  a 

certain  fraction  of  the  depth,  and  the  maximum  tensile  and  compressive 

intensities  of  stress  are  in  the  ratio  of  2  to  5.     Show  that  the  greatest 

economy  of  material  is  realized  when  theareasof  the  bottom  flange,  web, 

and  top  flange  are  in  the  ratio  of  25  to  20  to  4,  and  that  the  moment  of 

resistance  to  bending  is  \fAd,  where/=  Y"  maximum  tensile  intensity 

of  stress. 

• 

98.  Apply  the  results  in  the  preceding  question  to  determine  the  di- 
mensions of  a  cast-iron  beam  at  a  section  whose  moment  of  resistance  is 
800  in.-tons  and  whose  depth  is  18  in.,  taking  2  tons  per  square  incii  as 
the  maximum  tensile  mtensity  of  stress. 

Ans.  Ill  =  ^'-  sq.  in.;  A'  =  Vi"-  sq.  in.;  at  =  ff  sq.  in. 

99.  Determine  suitable  dimensions  for  a  cast-iron  girder  of  20  ft.  span 
and  24  in.  deep,  carrying  a  load  of  30,000  lbs.  at  the  centre,  the 
coefficients  of  working  strength  in  tension  and  compression  being  respec- 
tively 2000  and  5000  lbs.  per  square  inch. 

Ans.  fli  =  ijf^  sq.  in.  ;  A'  =  ^f-  sq.  in.  ;  a,  =  -4^  sq.  in. 

100.  A  cast-iron  girder  of  25  ft.  span  has  a  bottom  flange  of  36  sq.  in. 
sectional  area.  Find  the  most  economic  arrangement  of  material  for  the 
web  and  top  flange  which  will  enable  the  beam  to  carry  a  load,  of  18,900 
lbs.  at  10  ft.  from  one  end. 


dM^ 


EXAMPLES. 


423 


Ans,  Depth  =  20J  in.;  area  of  web  =  28.8  sq.  in.;  area  of  top 
flange  =  5.76  sq.  in. 
loi.  A  double-flanged  cast-iron  girder  lias  a  sectional  area  of  93  sq. 
in.;  the  web  is  i  in.  thici<  and  21  in.  deep;  the  moment  of  resistance  of 
the  section  is  100,950  ft. -lbs.  ;  the  coeHicients  of  strength  are  2100  lbs. 
per  square  inch  in  tension  and  5250  lbs.  in  compression.  Find  the 
position  of  the  neutral  axis  and  the  areas  of  the  two  flanges. 

102.  Deternune  the  moment  of  resistance  to  bending  of  a  section 
of  a  beam  in  which  the  to[)  flange  is  composed  of  f:i)o  340-min.  x  12-min. 
plates  and  one  340-mm.  x  lo-min.  plate,  aiul  the  bcjttom  flange  of  one 
j4o-nnn.  x  lo-mm.  [ilate  and  one  340-mm.  x  8-mni.  plate,  the  flanges 
being  riveted  to  a  1.4-m.  x  7-mni.  web  plate  by  means  of  four 
100-nim.  X  loo-mm.  X  8-mm  angle-irons.  The  coeflicient  of  strength 
=  6  k.  per  mm.'. 

103.  Compare  the  moments  of  resistance  to  bending  of  the  section  in 
the  preceding  question  and  of  a  section  in  which //^r^<^  400-mm.  x  1 5-mm. 
plates  are  substituted  for  the  top  flange,  and  one  400-mm.  x  15-mm, 
plate  is  substituted  for  the  bottom  flange. 

104.  Floor-beams  4.4m.  between  bearings  and  spaced  2.548  m.  c.  to  c. 
have  a  section  composed  of  two  equal  flanges,  each  consisting  of  two 
85-mm.  X  85-mm.  X  i2-mm.  angle-irons  riveted  to  a  490-mm.  x  7-mm. 
web.  A  weight  of  1 50  k.  (due  to  longitudinals;  and  a  weight  of  1 50  k.  (due 
to  rails, etc.),  i.e.,  300k.  in  all,  are  concentrated  at  the  rail-crossings,  and 
the  ties  have  also  to  carry  a  uniformly  distributed  load  of  400  k.  due  to 
weight  of  floor-beam,  4000  k.  due  to  weight  of  platform,  and  4000  k.  per 
square  metre  of  platform  due  to  proo/-\odid.  Find  the  moment  of  resist- 
ance to  bending  and  the  maximum  flange  intensities  of  stress. 

Ans.   /=  .000438584615. 

105.  The  section  of  a  beam  is  in  the  form  of  an  isosceles  triangle 
with  its  base  horizontal.  Show  that  the  momant  of  resistance  to 
bending  of  the  strongest  trapezoidal  beam  that  can  be  cut  from  it  is 
very  nearly  -^fbtP,  b  being  the  width  of  the  base  and  d  the  depth  of 
the  triangle. 

io6.  Taking/(,/c  as  the  tensile  and  compressive  intensities  of  stress, 
find  the  moment  of  resistance  to  bending  of  a  section  consisting  of  a 
2o/-in.  X  7/-in.  top  flange,  an  8o/-in.  x  io/-in.  bottom  flange,  and  a 
trapezoidal  web  4/  in.  thick  at  the  top,  8/  in.  thick  at  the  bottom,  and 
120/  in.  deep.  Also  compare  the  maximum  and  average  intensities  of 
shear. 

107.  Each  of  the  flanges  of  a  girder  is  a  350-mm.  by  lo-mm.  plate  and 
is  riveted  to  a  i.8-m.  by  8-mm.  web  by  means  of  two  loo-mm.  by 
loo-mm.  by  i2-mm.  angle-irons.  Determine  the  moment  of  resistance 
to  bending,  the  coefficient  of  strength  being  6  k.  per  square  millimetre, 


.-(.fTTir 


424 


THEORY  OF  STRUCTURES. 


n 


I .  M 


(a)  disiegarding  the  weakening  effect  of  riveting;  (b)  assuming  that  ihe 
flange-plates  are  riveted  to  the  angles  hy  20-inin.  rivets. 

Alls,  {a)  108661.04  km. 

108.  The  cross-tie  for  a  single-track  bridge  is  4. 1  m.  between  bearintjs, 
the  gauge  of  the  rails  being  1.5 1  ni. ;  each  of  the  flanges  is  composed  of  a 
148-iMni.  by  8-mm.  plate  riveted  to  a  550-mni.  by  8-intn.  web  by  means  01 
two  70-nim.  by  70-mm.  by  9-mm.  angle-irons;  a  hjad  of  296  k.  (weight 
of  rails,  etc.)  is  concentrated  at  each  rail-cro.ssiiig.  What  uiiifoniu'. 
distributed  load  will  the  tie  safely  bear,  the  metal's  coefficient  of  strenj^tli 
being  6  k.  per  square  millimetre  }  The  load  actually  distributed  over  tlic 
tie  is  19782  k.     Find  the  maximum  intensity  of  stress. 

Ans.  24162  k.  ;  4.94  k.  per  sq.  mm. 

109.  Design  a  longitudinal  of  .45  m.  depth  which  is  to  be  supported 
at  intervals  of  3.3  m.  and  to  carry  at  its  middle  point  a  weight  of  7000  k.. 
the  coefTicicnt  of  strength  being  5  k.  per  square  millimeter. 

Ans.  /=  259.875,  and  the  /of  a  section  with  two  equal  flan^'es, 
each  composed  of  two  70-mm.  by  70-mm.  by  9-mm.  angle-irons 
riveted  to  a  450-mm.  by  8-mm.  web  is  259. 102455. 

lie.  Find  the  moment  of  resistance  of  a  section  composed  of  two 
equal  flanges,  each  consisting  of  two  6oo-mm.  x  7-nim.  plates  riveted 
to  a  i20o-mm.  x  8-mm.  web  plate  by  means  of  two  loo-mm.  x  loo-rnni. 
X  12-mm,  angle-irons;  two  70-mm.  x  70-mm.  x  9-mm.  angles  are  also 
riveted  to  the  lower  faces  of  the  flanges,  the  ends  of  the  horizontal  arms 
being  24  mm.  from  the  outside  edges  of  the  flanges;  the  total  depth  of 
the  section  =  3.228  m.,  and  the  interval  between  the  two  web  plates, 
which  is  open,  is  2  m.;  coefficient  of  strength  =  6  k.  per  mm.". 

Ans.  /=  .093929232444  and  moment  =  349179.3018  km. 

111.  A  longitudinal  2.548  m.  between  bearings  consists  of  two  equal 
flanges,  each  composed  of  two  70-mm.  x  70-mm.  x  9-mm.  angle-irons 
riveted  to  a  350-mm.  x  7-mm.  web  plate.  Find  the  flange  intensity  of 
stress  under  a  maxinuim  load  of  7000  k.  at  the  centre. 

Ans.  /=  ,000139284508;  stress  =  5.6  k.  per  mm.". 

1 1 2.  A  cross-tie  resting  upon  supports  at  the  ends  and  2.26  m.  between 
bearings  is  composed  of  two  equal  flanges,  consisting  of  two  70-nini. 
X  70-mm.  X  9-mm.  angle-irons  riveted  at  the  top  to  a  450-mm.  x  7-!' 
web    plate    and  at  the  bottom  to  a  300-mni.  x  7-mm.  web   plate, 
interval  between  the  web  plates,  which  is  open,  being  2.55  m.  ;  the  ti 
designed  to  carry  a  uniformly  distributed  load  of  676  k.  per  lineal  mei: 
of  its  length,  and  also  a  load  of  1 1644.8  k,  at  each  of  the  points  distant 
.375  m.  from  the  bearing.     Find  the  position  of  the  neutral  axis  and  the 
maximum  flange  stresses. 

Ans.  1.516  m.  from  top  flange;    /=  .023194564198;    maximum 
.   .  B.  M.  =  481 5.8161  km.  ;  maximum  tensile  stress  =  .37  k.  per 

.    :    1       mm." ;  maximum  compressive  stress  =  .314  k.  per  mm.'. 


1  !■ 


EXAMPLES. 


425 


113.  Find  the  maximum  roiiccnlratcd  load  on  a  cross-tie  for  a  single 
track  due  to  a  six-wheel  locomotive,  the  wheels  being  2.3  m.  centre  to 
centre,  the  ties  being  3.2  m.  centre  to  centre,  and  the  weight  on  each 
wiicel  being  7000  k.  Ans.  10937.5  k. 

114.  The  floor-beams  f(jr  a  flouble-track  bridge  are  8.3  m.  betwet.'n 
hearings  and  are  spaced  2. 58  ni.  centre  to  centre.  The  distance,  centre 
to  centre,  between  track-rails  is  1.5  in.,  and  between  inside  rails  is  2  m.  ; 
liie  tie  has  equal  flanges,  each  consisting  of  two  70-mm.  x  70-mm,  x  9-mm. 
iin^'le-irons  riveted  to  a  6oo-mm.  x  7-mn\.  web;  the  maximum  live 
load  uij(jn  the  tie  is  that  due  to  a  weight  of  7000  k.  upon  each  of  the  six 
wheels  of  two  locomotives,  the  wheels  being  2.4  m.  centre  to  centre.  If 
the  coetficiiMit  of  working  strength  is  5I  k.  per  square  millimetre,  what 
uni((jrnily  distributed  load  will  the  tie  carry? 

115.  Determine  the  safe  value  of  the  moment  of  inertia  (/)  of  a 
cross-tie  for  a  double-track  bridge  ;  the  length  of  the  tie  between  bearings 
being  7.624  m.,  its  depth  .6  m.,the  gauge  of  the  rails  1.5  m.,  the  distance 
between  inside  rails  2  m.  The  uniformly  distributed  load  upon  a  tie 
consists  of  850  k.  per  square  metre  due  to  platform,  etc.,  and  of  1800  k. 
due  to  weight  of  tie  ;  the  ties  are  3.584  m.  centre  to  centre  ;  the  live  load 
is  that  due  t(j  a  weight  of  7000  k.  upon  each  of  the  centre  wheels  rjf  a  six- 
wheel  locomotive  and  a  weight  of  6000  k,  upon  each  of  the  front  and  rear 
wheels,  the  wheels  being  2.4  m.  centre  to  centre;  the  safe  coefficient  of 
strength  =;  6  k.  per  square  millimetre. 

116.  Tlie  upper  chord  of  a  Howe  truss  is  24  in.  wide  x  12  in.  deep 
and  is  made  up  of  four  12-in.  x  6-in.  timbers;  the  lower  chord  is  24  in. 
wide  X  16  in.  deep  and  is  made  up  of  four  i6-in.  x  6-in.  timbers  ;  the 
distance  between  the  inner  faces  of  the  chords  is  24  ft.  Find  the  mo- 
ment of  resistance  to  bending,  taking  800  lbs.  per  square  inch  as  the  co- 
efficient of  tensile  strength,  and  neglecting  the  effect  of  the  web. 

Ans.  Neutral  axis  is  137^  in.  from  bottom  face  of  lower  chord  ; 
moment  —  87441616  in. -lbs. 

Thr  cross-ties  of  a  single-track  bridge  consist  of  two  equal 
flani;cv,  eacii  comp<~oed  of  two  70-mm.  x  70  mm.  x  9  mm.  angle-irons 
riveter'  to  a  650-mm.  x  7-mm.  web;  the  ties  are  4.1  m.  long,  and  each 
carrii  9.146  k.  (viz.,  384  k./or //Vi-,  2762k.for  platform,  and  16,000  k.  for 
proof  load)  uniformly  distributed  and  635  k.  (due  to  longitiidinah,  rails, 
etc.)  concentrated  at  each  rail-crossing,  i.e.,  at  755  mm.  from  the  middle 
point.  Asscming  that  the  cross-ties  are  merely  supported  at  the  ends, 
find  the  maximum      tensity  of  stress. 

Ans,  5.7724  k.  per  mm." ;  -=.0018423. 

c 

N.B. — The  tixture  of  the  ends  approximately  doubles  the  strength. 


'in 


-  '-^l 


i'sH 

HI 

1 

426 


THEORY  OF  STRUCTURES. 


118.  The  lonj^itudinals  of  liic   bridge  in  the  last  question  consist  of 

two  i)airs  of   70-mm.      70-nun.  x  9-mni.   angle-irons   riveted  to  a  4  m. 

X  7  mm.  web;  the  cross-ties  are  3.2  m.  centre  to  centre.     Determine ilie 

niaximuni  intensity  of  stress  due  to  a  load  of  7000  k.  concentrated  on  the 

longitudinal  half-way  between  the  cross-ties,  assuming  that  it  is  an  iiiclr 

pendent  girder.     What  would  the  stress  be  if  the  ties  were  3  m.ceniic  to 

centre  ?  / 

Ahs.  -  =  .00095458  ;  5,866  k.  per  mm,";  5.4994  k.  per  mm,' 
c 

119.  The  section  for  the  Estressol  bridge  cross-ties  is  the  same  as  that 
for  the  Grande  Baise  (Ex.  1 17)  bridge  ties  ;  the  load  at  each  rail-crossin^f 
is  335  k.,  and  the  uniformly  distributed  load  is  18,062  k.  Find  tlie  max- 
imum intensity  of  stress  in  the  flanges,  assuming  that  the  tics  arc 
merely  supported  at  the  ends.  Ans.  5.26  k.  per  mm,' 

1 20.  In  a  rolled  joist  the  sum  of  the  two  flange  areas  and  tlic  nob 
area  is  a  constant  quantity.  Find  the  proportion  between  tluin  which 
will  give'a  joist  of  maximum  strength,  the  tliickness  of  the  web  i)eiiig 
fixed  by  practical  considerations.  Aits.  F'lange  area  =  8^  web  area. 

121.  An  aqueduct  for  a  span  of  20  ft.  consists  of  a  cast-iron  channel- 
beam  30  in.  wide  and  20  in.  deep.  Find  the  thickness  of  the  metal  so 
that  the  water  may  safely  rise  to  the  toj)  of  the  channel,  the  safe  coi'lli- 
cient  of  strength  being  i  ton  per  square  inch.  Find  the  safe  limiting 
span  of  the  channel  under  its  own  weight. 

122.  A  rolled  beam  with  equal  flanges  and  a  web  whose  section  is 
equal  to  the  joint  section  of  the  flanges  has  a  span  of  24  ft.  and  carries  a 
weight  of  8  tons  at  the  centre.  If  the  stiffness  is  .001  and  if  the  cocilicicMt 
of  strength  per  square  inch  is  5  tons,  find  the  depth  of  the  beam  and  the 
web  and  flange  sectional  areas. 

123.  A  wrought-iron  beam  of  'section,  20  ft.  between  supports, 
carries  a  uniformly  distributed  load  of  4<x)o  lbs.  and  deflects  .1  in.;  the 
eflfv-ctive  depth  =8  in.;  A"  =  30,000,000  ibs.;  web  area  =  joint  area  of 
the  equal  flanges.  Find  the  total  sectional  area.  Also  find  the  width  of 
a  rectangular  section  8  in.  deep  which  might  be  substituted  for  the 
above.  Ans.  I  —  288;  area  =  27  sq.  in.;  widtli  =  6!i  in. 

124.  A  cast-iron  beam  of  an  inverted  T-section  has  a  uniform  depth 
of  20  in.  and  is  22  ft.  between  supports;  the  flange  is  12  in.  wide  and  1.2 
in.  thick  ;  the  web  i"  i  in.  thick;  the  load  upon  the  beaiti  is  4500  lbs. 
per  lineal  foot;  /{"=  17,000,000  lbs.  Find  the  deflection  at  the  centre, 
the  moment  (/f  resistance  to  bending,  the  maximum  tensile  and  com- 
pressive intensities  of  stress,  and  the  position  of  the  neutral  axis.  Why 
is  the  flange  placed  downwards  ? 

125.  Find  the  sectional  are^x  of  a  wrought-iron  beam  of  T-section 
whicli  may  be  substituted  for  me  cast-iron  beam  in  the  preceding  qnes- 
tion.  the  depth  being  tiie  same  and    the   coefficients  of   strength   per 


n  consist  of 
I  to  .1  4  in. 
itcrminctlie 
•ated  oiiihc 
t  is  an  iiKJc 
m.  centre  id 

.  per  mm.' 

same  as  that 
■ail -crossing 
nd  tile  iiiax- 
the  tics  are 
per  mm.' 
nd  the  h'oIj 
them  wliicli 
web  being 
web  area. 

on  cliannel- 

lie  metal  so 

safe  cdi'iri- 

afc  limiting 

e  section  is 
nd  carries  a 
e  coeiruienl 
;am  and  the 

n   supports, 

s  .1  in.;  the 

lint  area  of 

he  width  of 

ted   for  the 

=  6'|  in. 

form  (lipth 

ide  and  i.: 

s  4500  lbs. 

the  centre, 

5  and  com- 

laxis.     Why 


EXAMPLES. 


427 


quare  inch  being  3  tons  in  compression  and  5  tons  in  tension.  W^iy 
should  the  flanjje  be  uppermost  ?  What  should  the  total  sectional  area 
be  if  the  flange  and  web  are  of  equal  area  ? 

126.  A  cast-iron  girder  139  in.  between  supports  and  10  in.  deep  had 
atop  flange  2i  in.  x  |  in., a  bottom  flange  10  in.  x  1}  in., and  a  web  }  in. 
tiiick.  The  girder  failed  under  loads  of  lyi  tons  placed  at  the  two 
points  distant  3f  ft.  from  each  support.  What  were  the  central  flange 
stresses  at  the  mompiit  of  ruptur;?  What  was  the  central  deflection 
wli'M)  tiie  load  at  each  point  was  7i  tons  ?  (£"  =  18,000,000  lbs.  ;  weight 
of  girder  =  3368  lbs.;  ton  =  2240  lbs.) 

Afis.  18225..';}  lbs.  =;  total   flange    stress;    unit  flange  stresses 
=  14,580,  and  41,657  lbs.  persq.  in.;  deflection  =  .35". 

127.  A  cylindrical  beam  of  2  in.  diameter,  60  in.  Ui  length,  and  weigh- 
ing ]  lb.  per  cubic  inch,  deflects  ^\  in.  under  a  weight  of  3000  lbs.  at  the 
ceniic.     Find  £".  Aus.  £'=21,645,511  lbs. 


I 


.ill- 


f  T-soction 
ediug  qnes- 
rcngtli   per 


n  M, 


:i. 


;t  V   , 


;!•  1    ::l:il 


CHAPTER   VII. 

ON  THE  TRANSVERSE  STRENGTH    OF    BEAMS.— Conhniud 

I.  General  Equations. — The  girder  OA  of  length  /  carrier 
a  load  of  which  tl^e  intensity  varies  continuously  and  is  />  at  a 
point  K  distant  x  from  O. 

M^    M+tIM 

/     \ 

'     K     V   K'* 


IV 


L   I     L' 
Fig.  3to. 

Consider  the  conditions  of  equilibrium  of  a  slice  of  the 
girder  bounded  by  the  vertical  planes  KL,  K'L',  of  which  the 
abscissae  are  x,  x  -{•  dx,  respectively. 

The  load  between  these  planes  may,  without  sensible  error, 
be  supposed  to  be  uniformly  distributed,  and  its  resultant  pdx 
therefore  acts  along  the  centre  line  W . 

The  forces  acting  upon  the  slice  at  the  plane  KL  are  equiv- 
alent to  an  upivard  shearing  force  S,  and  a  right-handed  couple 
of  which  the  moment  is  M,  while  the  forces  acting  upon  the 
slice  at  the  plane  K'L'  are  equivalent  to  a  dowmvard  shearing 
force  S  -\-  dS,  and  a  left-handed  couple  of  which  the  moment 
\i>M^dM. 

Since  there  is  to  be  equilibrium,  , 

S--{^S-\-dS)—pdx=^  the  algebraic  sum  of  the  vertical  forces =o. 

dS 


dx 


-\-p  =  O ((?) 

42S 


GENERAL   EQUATIONS. 


429 


And,  M-{M-^  dM)  +  S—  +  (5  +  ^5)  —  =  the   alge- 

braic  sum  of  the  moments  of  the  forces  with  respect  to  V  or 

V  =  o. 


dM      „ 

,  — 5  =  0. 

dx 


(^) 


f  ^     7 

The  term  — '- —  is  disregarded,  being  indefinitely  small  as 

compared  with  the  remaining  terms. 

Equations  {a)  and  {b)  are  the  general  equations  applicable 
to  girders  carrying  loads  of  which  the  intensity  is  constant  or 
varies  continuously.  Their  integration  is  easy,  and  introduces 
two  arbitrary  constants  which  are  to  be  determined  in  each 
particular  case. 

Cor.  I.  From  equations  {ci)  and  {b\ 

dx'  ~  dx~      ^' 

Let  p  =  wf{x\  w  being  a  constant,  and  f{x)  some  function 
of  X.    Then 

and 


M 


= f, + c,x  -  tv  r  rf{x)dx\ 


|l!  !■ 


f,  and  r.j  being  the  constants  of  integration,  and  o  and  x  the 
limits. 

Example. — Let  the  girder  rest   upon   two  supports    and 
cany  a  uniformly  distributed  load  of  intensity  w^.     Then 


dM 


am  px    J 

— — -  =  c^  —  I    iVydx  =  t,  —  w^x, 

dx  v% 


and 


M=  c^-\-  c,x  —  w, 


430  THEORY  OF  STRUCTURES. 

But  M  is  zero  when  ^  =  o  and  also  when  x  =  /.     Hence 


Therefore, 


and 


f ,  =  o    and     r, 


W.l  TV.      „ 

M=  -~-x -x\ 

2  2        ' 


^      dM      zv.l 

S  —  -y  =  —  ZV.X. 

dx  2  ' 


ik_._  il  ii 


Cor.  2.  The  bending  moment  is  a  maximum  at  the  point 

defined  by  -y-  =  o  =  S,  i.e.,  at  a  point  at  which  the  shearing 
dx 

force  vanishes. 

In  tlie  preceding  example,  the  position   of  tlie   maximum 

bendmg  moment  is  given  by  .S  =  o  =  -—  7i'^x,    or  x  =  -, 

and  its  correspondmg  value  is —  —  —  =  —^-. 

^  22248 

The  shearing  force  is  greatest  and  equal  to  —  when.T=o. 

Cor.  3.  Suppose  that  the  load,    instead  of  varying  contin- 
.,  M    , ,      uously,  consists  of  a  number  of  finite  wci'fhts 


at  isolated  points. 

By  reason  of  the  discontinuity  of  the  load- 

Fig.  311.  jpg^  ^.j^g  general  equations  can  only  be  inte- 

grated between  consecutive  points. 

Let  Nr,  ^r+i »  be  any  two  such  points,  of  abscissae  x,. ,  x^+,, 
respectively. 

Between  these  points  equations  {a)  and  (d)  become 


dS  ,     dM       ^ 

^  =  0,     and     ^;  =  5. 


.'.  S  =  a  constant  =  S,.,  suppose,  between  N^  and  N,.+, 


■  i  f  i- 


GENERAL  EQUATIONS. 


43 1 


dM 


Hence,   '^  =  S^,  and   M=  S^x  -\-  c,    between   N^  and 

ilX 

Nr^,  ,  c  being  a  constant  of  integration. 
Let  M  —  Mr  wlicn  x  =  x^-     Then 


C  =   Mr  —   SrXr  ,       and        M  =   Sr{x  —  Xr)  +  Mr  . 


maximum 


Also,  if  M  —  Mr+,  when  x  —  Xr^, , 


M..+  ,  =--  S,.(^.+  ,  -  Xr)  +  i/;. 


The  terminal  conditions  will  give  additional  equations,  by 
mc.'uis  of  which  the  solution  may  be  completed. 

Example. — The  girder  OA,  of  length  /,  rests  upon  two 
supports  at  O,  A,  and  carries  weights  /', ,  P, ,  at  points  B,  C, 


4 


R> 


R, 


Tx 


Fig.  312. 

dividing  the  girder  into  three  segments,  OB,  BC,  CA,  of  which 
the  lengths  are  r,  s,  t,  respectively. 

The  reaction  R,  a.\.  O  =  ^^^^+J)  +  ^^\ 
The  reaction  i?,  at  /J  =  ^^  +  ^'^(^  +  .y)  ^ 

Between  0  and  B,  S  is  constant  =  5,.  suppose,  =  /?, , 

.-.  M  =  SrX, 

there  being  no  constant  of  integration,  as  M=  o  when  x  =  o. 
Also,  when  x  =  r,  M  =  SrT. 


% 

H 

H! 

^  IT 

m 

J 

i 

!. 

\ 

=^JP^i» 


i^ii;;  I 


432  THEORY  OF  STRUCTURES. 

Between  B  and  C,  S  is  constant  =  S,    suppose,  =  /?,  —  /*,. 

c'  being  the  constant  of  integration. 
But  yl/=  S^T  when  x  —  /-. 

.'.  c'  —  {Sr  —  S^r,    and     M  =  S,x  -{-  {S^  —  S,)r. 

Also,  v.'hen  x  ■■=  r  -\-  s,  M  =  S^  -f-  S^r. 
Betwcien  6" and  A,  S  is  constant  =  S/  suppose,=y?,— /*,— P„ 
and  hence 

M=S,x-\-c", 

c"  being  tha  constant  of  integration. 
But  M  =  S,s  +  •S,f  wlien  x  =  r  -{- s. 


.'.  c"  =  S,s  -{-  S,r  —  St[r  -f-  s), 


and 


S'W 


M  =  V  +  S,s  4-  5,7-  -  S^{r  -\-  s). 


Hence,  zt  A,  o  =  5/  +  5^^  +  S^r 


Cor,  4.    The  equation 


dx 


S  indicates  that  the  shearing 


force  at  a  vertical  section  of  a  girder  is  the  increment  of  the 

bending  moment  at  that  section  per  unit  of  length,  and  is  an 

important  relation  in  calculating  the  number  of  rivets  required 

for  flange  and  web  connections. 

2.  On  the  Interpretation  of  the  General  Equations  — 

The  bending  moment  M  at  any  transverse  section  of  a  girder 

FI 
may   be  obtained  from  the   equation  M=  ~,R  being  the 


H  -J 


DEFLECTION   OF  BEAMS. 


433 


/?,  -  Pv 


:.)r. 


^'P-P. 


nent  of  the 
h,  and  is  an 
ets  required 

Equations.— 

of  a  girder 

V  being  the 


radius  of  curvature  of  the  neutral  axis  at  the  section  under 
consideration. 

Let  OAy  in  Figs.  313  and  314,  represent  a  portion  of  the 
neutral  axis  of  a  bent  girder. 

0.= 1 ^ X 


■wX 


Fig.  313 


Fig.  314. 


Take  O  as  the  origin,  the  horizontal  line  OX  as  the  axis  of 
.r, and  the  line  OY  drawn  vertically  downwards  as  the  axis  o{ y. 

Let  x,y  be  the  co-ordinates  of  any  point  P  in  the  neutral 
axis. 

If  R  is  the  radius  of  curvature  at  P,  then 


R=^ 


"dx" 


1-H©T 


..dd 
=  ±cos6^^; 


the  sign  being  -f-  or  —  according  as  the  girder  is  bent  as  in 

Fig-  313  or  as  in  Fig.  314,  and  B  being  the  angle  between  the 

tangent  at  P  and  OX. 

dy 
Now,  -^  is  the  tangent  of  the  angle  which  the  tangent  line 

at  /'to  the  neutral  axis  makes  with  OX,  and  the  angle  is  always 

^m  dy 

,       •  „    ■     very  small.     Thus,  ■—  is  also  very  small,  and   squares  and 
he  shearing    ■  -^  '  dx  j'  >  m 


dy 
higher  powers  of  -j-  may  be  disregarded  without  serious  error. 

Hence,  •    . 

^=±^i-,   approxnnately, 

and  the  bending-moment  equation  becomes 

d^v 


JllBfigJt, 


ii,:. 


hh 


1^ 


434 


TI/EOKV  OF  STRUCrUKKS. 


The  integration  of  this  equation  introihiccs  two  arhitran- 
constants,  of  whiclj  the  v.ihicsarc  to  he  dctrrniincd  from  j,'iv('ii 
conditions.  At  the  point  or  points  of  support,  for  exain[)lc, 
the  neutral  axis  may  bo  horixA)ntal  or  may  sU)pe  at  a  j^ivcii 
an^rlc. 

Let  0  he  the  slope  at  /'.     Since  0  is  <;enerally  very  small. 


aiul  hence 


W  =  tan  (K  api^roxiniately, 


or 


try      dB  M 

dx       ^  EI 


dx* 


(A) 


From  this  last  equation 

e  =  ±  j^/Mdx, 

and  the  c/untj^i'  of  slope  between  any  given  limits  is  represented 
by  the  corresponding  area  of  the  bending-moment  curve. 


Also,  since    "-  =  ^,       y  ~  J  ^''''*'» 


and  the  deflection  is  measured  by  the  area  of  a  curve  repre- 
senting the  slope  at  each  point. 
Again,  by  Art.  i, 


d'M     dS 
dx*  ~  dx 


=  -p. 


(B) 


Comparing  eqs.  (A)  and  (B),  it  will  be  observed  that 
y,  a,  and  TjT^.,  i.e.,  the  deflection,  slope,  and  bending  moment, 
are  connected  with  one  another  in  precisely  the  same  manner 


NEVTh'AI.   AXIS  01'    A    tOADED   liEAM. 


435 


s 


as  My  5,  and  p,  i.e.,  the  liciidiiifj  moment,  shearin^^  force,  and 
lo.id.  Tltiis,  llu:  mutual  relations  hi-tween  curves  drawn  to 
iipri'si-nt  tlie  tifjlrction,  slope,  .umI  /uiu/inx  mouicut  must  he  the 
same,  mutatis  mutandis,  as  those  between  the  turves  of  bendin^f 
inoMient,  slu-arin^  force,  and  load. 

l''or  example,  ilivide  the  ijD'cctivc  bending-moment  area  hito 
a  number  of  elrmentary  areas  by  drawing  vertical  lines  at  con- 
venient distances  apart,  and  su|)pose  these  elementary  areas  to 
irprc.'sent  weights.  Two  reciprocal  figures  connecting  j/,  ^,  and 
;)/  may  now  be  drawn  exactly  as  described  in  Chap.  I,  and  it 
at  once  follows  that — 

(rt)  Any  two  sides  of  the  funicular  polygon,  or,  in  the  limit 
(when  the  widths  of  the  elementary  areas  are  indefim'tely 
(liniinished),  any  two  tangents  to  the  funicular  or  dcjlection 
curve,  meet  in  a  point  which  is  vertically  bilow  the  centre  of 
gravity  of  the  corresponding  ijj'cctivf  moment  area. 

(/;)  The  segments  !//,  ////into  which  the  line  of  weights  is 
divided  hy  drawing  OH  parallel  to  the  closing  line  CD,  give  the 
slopes  (=  ^Mdx)  at  the  supports. 

N.B. — In  the  casti  of  a  semi-girder,  the  last  side  of  a 
polygon  is  the  closing  line,  and  \n  gives  the  total  change  of 
>l()pe. 

{c)  If  the  polar  distance  is  made  equal  to  EI,  the  intercept 
between  the  closing  line  and  the  funicular  or  deflection  curve 
measures  the  deflection. 

3.  Examples  of  the  Form  assumed  by  the  Neutral 
Axis  of  a  Loaded  Beam. 

ExAMi'i.K  I.  A  semi-girder  fixed  at  one  end  O  so  that  the 
neutral  axis  at  that  point  is  horizontal  carries  a  weight  W  at 
tlie  other  end  A.     At  any  point  {x,  y)  of  the  neutral  axis 

-^^Elf^,  =  W{l~x).   .     .     (A) 
Integrating, 

^,  being  a  constant  of  integration.    But  ^'°'  3«s- 

tiic  girder  is  fixed  at  O,  so  that  the  inclination  of  the  neutral 


!    I  I 


1:  '• 


I 


436 


THEORY  OF  STRUCTURES, 


axis  to  the  horizon  at  this  point  is  zero,  and  thus,  when  ;r  =  0, 

dy 

■J-  is  o,  and  therefore  c.=^o. 

ax  ' 

Hence,  ' 


EI%^W{^.-^ 


(B) 


Integrating, 


£/,=  (f(/?'-^;")+.„ 


f,  being  a  constant  of  integration.      But  y  =  0  when  x  —  0, 
and  therefore  c,  =  o. 
Hence, 


E^y  =  iv[i'^ -^) 


(C) 


dy 


Equation  (B)  gives  the  value  of -^  ,  i.e.,  the  slope,  at  any 

point  of  which  the  abscissa  is  ;f. 

Equation  (C)  defines  the  curve  assumed  by  the  neutral  axis, 
and  gives  the  value  of  y,  i.e.,  the  deflection,  corresponding  to 
any  abscissa  x. 

Let  or,  be  the  slope,  and  d^  the  deflection  at  A. 

From  (B), 


I  wr 


and  from  (C), 


d,  = 


3  £  r 


NEUTRAL  AXIS  OF  A   LOADED  BEAM. 


437 


Ex.  2,  A  semi-girder  fixeJ  at  one  end  0  carries  a  uniformly 
distributed  load  of  intensity  xv. 

At   any   point   P  {x,  y)    of    the  j 
neutral  axis, 


,(Vy      XV 


A-EI-.-;^^    M-xf 


dx'         2 


W . 


=  -    (l'  -2/x+x').    .     (A) 

2  ^  '         ''  ^     '  Fio.  3i«.     . 

Integrating, 

t-,  being  a  constant  of  integration. 

dy 
But  -,—  =  o  when  x  =  o,  and  therefore  f,  =  O.     Hence, 

■^^%=jh-'^'+i) m 

Integrating, 

6',  being  a  constant  of  integration. 

But  ^  =  o  when  x  =  o,  and  therefore  f,  =  o.     Hence, 


('T-4+f^) (^) 


W  I     X 
-^         2  \     2  3         12 


Let  ff,  be  the  slope  and  d^  the  deflection  at  A.     Hence,  from 


iB), 


and  from  (C), 


I  wP 
tana,  =  g^; 


_  I  wl* 


i    ■' 


438 


THEORY  OF  STRUCTURES. 


Ex.  3.  A  semi-girder  fixed  at  one  end  carries  a  uniformly 
distributed  load  of  intensity  w,  and  also  a  single  weight  WdX  the 
free  end.  Tiiis  is  merely  a  combination  of  Examples  I  and  2, 
and  the  resulting  equations  are  : 

El''^^W{l-x)Jr'^-(l-x)' (A) 

Also,  if  A  is  the  slope  and  D  the  deflection  at  the  free  end, — 
from  (B), 

tan  A  =  Yl\~r  "^  ~^)  "^  *^"  "'  +  *^"  "' ' 
and  from  (C), 


I  (wr  ,  w/*\ 


£l\  3 

Cor. — The  slope  (a)  and  deflection  (d)  of  an  arbitrarily 
loaded  semi-girder  may  be  determined  in  the  manner  de- 
scribed in  Art.  2. 

Let  F  be  the  area  of  the  bending-moment  curve.  Its 
centre  of  gravity  is  at  the  same  horizontal  distance  .*•  from  the 
vertical  through  A  as  the  point  T  in  which  the  tangent  at  A 
intersects  OX. 


.'.  -F>=  a  =  angle  A  TX  =  -. 


In  Ex.  3,  e.g., 


£1 


2    '    3     2    ' 


and 


Fx 


wr2      wP% 

i+-^-i=Erd. 

2364 


NEUTRAL  AXIS   OF  A    LOADED   BEAM. 


439 


Note. — If  the  semi-girder  in  the  three  preceding  examples 

is  on\y  partially  fixed  at  O,  so  tiiat  the  neutral  axis,  instead  of 

being  horizontal  at  the  support,  slopes  at  an  angle  ^,  then 

dy 
when  X  ^=o,  -.—  =  tan  &,  and  the  constant  of  integration, 

c, ,  is  also  ^/tan  ^.     Thus,  the  left-hand  side  of  eqs.  (B)  and 
(C),  respectively,  become 

£f[j-  -  tan  ^j     and     £/(j  -  x  tan  ^). 


Ex.    4.     The    girder  OA    rests   upon   two  supports   at    O, 


A,  and  carries  a  weight  W  at  the 
centre. 

The  neutral  axis  is  evidently 
symmetrical  with  respect  to  the 
middle  point  C,  and  at  any  point 
P  {x,  y)  between  (?  and  C, 


0 


B 


F^ 


p  -f 

w 

Fig.  317. 


Integrating, 


dx         2 


dx        4         '     * 


....     (A) 


£,  being  a  constant  of  integration. 

But  the  tangent  to  the  neutral  axis  at  C  must  be  horizontal, 

so  that  when  x  —  -,  -y-  =  O,  and  therefore  c.  = 2~  • 

2   dx  '  16 


Hence, 


Integrating, 


—  EI-T-  =  —x^ ^ 

dx        4  ID 


(B) 


vr        w  ,     ivr    ^ 

Ely    =-x  --^^  +  ,., 


c,  being  a  constant  of  integration. 


* 


:  I. 


';m 


440  THEORY  OF  STA'UCTUJfES. 

But  J  =  o  when  x  =  o,  and  therefore  f,  =  o.     Hence 


-  Efy 


w  ,     wr 

12  16 


(C) 


Cor. — Let  «,  be  the  slope  at  0,  and  d,  the  doflection  at  the 
centre.     Then, 

I  wr  I  WP 

from  (B),  tan  a,  =  j^  -^ ;  and  from  (C),  d,  =  ~  ~pj. 


B 
P  ~c" 


Fig.  318. 


^  Ex.    5.    The    girder    OA    rests 

'  '^  upon  supports  at  O,  A,  and  carries 
a  uni'"ormly  distributed  load  of  in- 
tensify tu. 

Ai   any  point  /'  {x,  j>)    of  the 
neutral  axis, 


integrating, 


„  -  dy       xvl  TVX' 

EI~  —  — X 

dx         2  2 


dy       wl  zvx^    , 

dx        4  6      '     " 


(A) 


I- 


t: 


m 


c^  being  a  constant  of  integration. 


dj 


I 


But  -7-  =  o  when  x  —  -,  and  therefore  tr,  =  — 
dx  2  ' 

Hence, 


24 


^j^      w/  ,      wx^      wP 
—  EI  -J-  =  — ,r' ^ •.       .    ,     .    (B) 

dx         A  6  24  •      •      V    / 


Integrating, 


24 

wx^    wr 


wl    ^       ..^ 

Ely     =  —A-' X  4-  <:,, 

12  24        24      '     • 


c,  being  a  constant  of  integration. 

But  y  -—Q  when  x  =  o,  and  therefore  r,  =  o. 
Hence 

-  Ely  =  —.1-' ^^  -  -~x (C) 


12 


24        24 


NEUTRAL   AXIS  OF  A    LOADED   BEAM. 


441 


Let  <*,  be  the  slope  at  0,  and  d.^  the  deflection  ;it  the  centre, 
Tlien, 

I  ivl^  c    wl^ 

from  (B),  tan  ix.  = ^,--7- ;  and  from  (C),  d.  z=l  — -    ----. 

^    '  24,hl  \    /'    1       2,^4  hi 

Ex.  6.  A  girder  rests  upon  two  supports  and  carries  a  uni- 
formly  distributed  load  of  intensity  w,  together  witii  a  single 
woif^iit  VV  at  the  centre.  This  is  merely  a  combination  of 
Examples  4  and  5,  and  the  resulting  equations  are : 


nl'r.1  =  — X  H •^ 

dx         2        '     2  2 


(A) 


and 


dv        W  W         wl  wx^      wr 

-El-f   =-x'  -  ~-P  J^—x-'  -  -^  -^^  ,    .    (B) 
dx         4  16       '     4  6         24  ^    ' 


W         W 

-  Ely       =  -x'  -  -J'x  +  -x' X.     (C) 

•^  12  16        '    12  2/         "^  -^ 


7C'/   .         Tt'Jir*        w/' 

24        24 


Also,  if  A   is  the  slope  at  the  origin,  and  D  the  central 
deflection,  we  have,  from  (B), 


tan/J  =  -^l^-^  +  -— -j  =  tan  a,+  tan  a,; 


and  from  (C), 


^=ii{'^+h"'')  =  <+<- 


Cor. — The  slope  and  deflection  of  an  arbitrarily  loaded 
t,nrdcr  resting  upon  two  supports  may  be  determined  in  the 
manner  described  in  Art.  2. 

Let  C  be  the  lowest  point  of  the  deflection  curve.  The 
tan<^cnts  at  C  and  0  will  intersect  in  a  point  T  which  is  ver- 
tically below  the  centre  of  gravity  of  the  bending-moment 
area  corresponding  to  OC. 

Denote  this  area  by  F  and  the  hoi  izontal  distance  of  centre 


442 


THEORY  OF  STRUCTURES. 


of  gravity  from  OY  by  ^.     Let  ix  be  the  angle  between  01' 
and  CT  produced.    Then 

EI 


a 


^  being  the  maximum  deflection. 

In  Ex.  6,  e.g.,  the  girder  being  synnnetrically  loaded, 


I.  f 


242'382  i6  32'    24  82 


m7\' 

0 


Ex.  7.  Suppose  that  the  end   O  of  the  girder  in   Ex.  5 

_  „         is  fixed.     The   fixture  introdiicfs 

^R,  Ra 

I 


4 -'"A 

P 


Fig.  319. 

neutral  axis, 


a  Icft-liaiidcd  couple  at  0 ;  let  its 
■X  moment  be  J/,. 

Let  the  reactions  at  O  and  A 
bo  R^ ,  A', ,  respectively. 

At  any  point  /'  {x,  y)  of  the 


-  £7-7-4  =  ^,'f ^, 


(I) 


'*) 


But  M,  i.e.,  —  £I-j^>  is  zero  when  x  =  /, 


\hi 


7x>r 

...  o  =  A,/ J/., 

1  ^  1 


Integrating  eq.  (i), 


^rdy       „x^      wx' 

dy 


(2) 


(3) 


There  is  no  constant  of  integration,  as  -,-  =  o  when  x  =  0. 
Integrating  eq.  (3), 


-  E/y  =  R, 


x*      zvx*        .,x* 


24 


M,-. 


(4) 


There  is  no  constant  of  integration,  as  x  and  y  vanish  to- 
gether. 


m 


cd, 

I 

2  ~ 

:zEId. 

in 

Ex.  5 

introduces 

0; 

let  its 

\.0 

aiul  /) 

.y) 

of  the 

• 

.     (1) 

NEUTRAL  AXIS  OF  A   LOADFD  BEAM. 
But  7  also  vanishes  when  ;r  =  /,  so  that 

o^'        '^i'         n^f 

0  =  R,2- MJ.  .    .    .    , 

'6        24  ' 

Hence,  by  cqs.  (2)  and  (5), 


445 


(5> 


M,  =    g-,     A',  ==  |7f/,     and  so    R^  -^  |w/.      .     (6) 

Thus,  the   bcnding-momcnt ,  slope,  and   deflection  equations 
arc,  respectively, 

7i'  ,,     IV r 
2'  --jt 


M, 


Mr  16 


W  7VP 

6'  -  ^^'^' 


5  w/ 

■^  48  24 


7VP 
16 


,-.r 


(7) 
(8) 
(9) 


C(3r.  I.  The  bending  moment  is  «//at  points  given  by 

5    ,         «^  .      w^ 


i.e.,  when  x  ^=  -  or  I.     Take  (?/^  =  — .     K 
4  4        \ 

Since  y-7  =  o,  /^  is  a  point  of  inflexion. 


V 


,:^' 


^A 


If  the  girder  is  cut  through  at  this  point, 

and   a    hinge    introduced    sufficiently   fi*" 

strong  to  transmit   the  shear  {=  ^ivl),  Fio.  320. 

tlic  s, ability  of  the  girder  will  not  be  impaired. 

Hence,  the  girder  may  be  considered  as  made  up  of  two 
independent  portions,  viz. : 

/ 

{a)  A  cantilever  OF  oi  length  — ,  carrying  a  uniformly  dis- 

4 

trihuted  load  of  intensity  w,  together  with  a  weight  \zul  at  /*". 


444 


THEORY  OF  STRUCTURE S. 


The  maximum  bending  moment  on   OF  is  at  0,  and  is 

3     .  I   .  tvl  I      xvP 


^s^^i  +  Ts^X" 


ii 


(/;)  A  girder  FA  of  length    -,  carrying  a  uniformly  distrib- 
uted load  of  intensity  re'. 

The  maximum  bending  moment  on  FA  is  at  the  middle 

8  W  >'  ~  128' 
This  result  may  also  be  obtained  from  eq.  (7)  by  putting 

-— -  ■=  o.     Whence 
dx 


point  D,  and  is  —  -  l '  / )   = 


O  =  ^zvl  —  zvx,     or     X  =  §/, 


and  therefore 


M., 


tI¥«'^° 


The  shearing  force  and  bending  moment  at  different  points 
of  the  girder  may  be  represented  graphical!)  as  follows  : 

The  shearing  force  at  any  point  of  which  the  abscissa  is  x  is 

5  =  %xvl  —  tvx. 

Take  OB  and  AC,  respectively  equal  or  proi)crtional  to  i;ri' 
and  fW;  join  BC  The  line  BC  cuts  OA  in  B,  where  OD  —  |r 
The  shearing  force  at  any  point  is  represented  by  the  ordinate 
between  that  point  and  the  line  BC. 

The  bending  moment  at  the  point  {x,jy)  is 

5  7CI  ivl* 

M  =  k'^vIx x'' 5-. 

828 

Take  OG,  DE,  and  OF,  respectively  equal  or  proportional 

W'      9      ,,         ,  ^      0-.     .       ,• 
to  -5—,     ^uu\  and  -.      Ine  benduig  moment  at  any  point  is 
o       128  4 

represented  by  the  ordinate  between  that  point  and  the  parab- 
ola passing  through  G,  F,  and  A,  having  its  vertex  at  A  .nii' 
its  axis  vertical. 


NEUTRAL  AXIS  OF  A   LOADED   BEAM. 


dy 


44S 


Cor.  2.  The  deflection  is  a  maximum  wiien  -.-  =  o,  i.e.,  when 

ax 


wl' 


5  7C 

i6  6  8 


or  at  the  point  ^iven  by  jr  =    5 ('5  —  ^^ii)- 

Substituting  this  value  of  x  in  cq.  (9),  the  corresponding 

value  of  y  may  lie  obtained. 

Ex.  8.  If  both  ends  of  the  girder  in  eq.  (7)  are  fixed,  the 

wl 
reaction  at  eacli  support  is  evidently  — ,  and  the  equation  of 


moments  becomes 


—  LI--/:,  —  '-X M, 

dx'        2  2  ' 


(I) 


Integrating, 


-  EI%  ^  '^x'  -  Ix^ 
dx        4  6 


M.x. 


dy 


.     .      .     (2> 


No  constant  of  integration  is  required,  as  -7-  —  o  when  x  =.  o. 

;  -  is  also  zero  when  x  r=  I  (also  when  x  =  -]. 
dx  \  2  / 


wl*      wl* 
4  6 


dud  hence 


M,= 


xvl* 
12 


Integrating  eq.  (2), 


wl   J,       zv 


wP 


—  Ely  =  —  X X' X' 


(.3) 


(4) 


24  24 

There  is  no  constant  of  integration,  as  x  and  j' vanish  together. 

1  he  central  deflection  I  i.e.,  when  x  =  --  j  =  — ,    -y^' 

\  2  /       384  El 


446 


THEORY  OF  STRUCTURES 


If  tlie  load,  instead  of  being  uniformly  distributed,  is  a 
weight  (F  concentrated  at  the  centre,  then,  {ox  one  //^z//"  of  the 
girder, 


dx'        2  ' 


(5' 


N- 


^(4  ;:, 


Integrating, 


dy        W 
EI-^=  —x-'-M.x. 
dx        4  ' 

dy 


(6, 


There  is  no  constant  of  integration,  as  -t-  =  o  when  x  —  O). 


dy 


dx 


7  is  also  evidently  zero  when  x  =  -,  and  hence 


i6  '2  '8 


•    (7) 


Integrating  eq.  (6), 


W  x" 

ETv  =  —X'  -  M, 

12  ' 


w 


Wl 


.    .    (8) 


2  12  l6       ' 

There  is  no  constant  of  integration,  as  x  and  y  vanisli  together. 

I     Wi 


The  central  deflection  = 


192  EI' 


4.  Supports   not   in   same   Horizontal   Plane. — In  the 

preceding  examples  it  has  been  assumed  that  the  ends  of  the 
girder  are  in  the  same  horizontal  plane.  Suppose  that  one  end, 
e.g.,  A,  falls  below  0  by  an  amount  jk,  ,  y^  being  small  as  com- 
pared with  /. 

The  abscissae  of  points  in  the  neutral  axis  are  not  sensibly 
changed,  but  the  conditions  of  integration  are  altered.  Con- 
sider Ex.  4. 

Between  0  and  C, 


(>) 


ited,  is  a 
alf  of  the 

.     .    (5' 


.     .    (6' 


tien  ;f  =  0. 


(7> 


.    .    .    (8> 
|sh  together. 


le.— In  the 

lends  of  the 

lliat  one  end, 

lall  as  com- 

lot  sensibly 
lered.     Con- 


.    ('^ 


447 


^/^=  ---V  +<^> (2) 


NEUTRAL  AXIS  01-   A    LOADED  BEAM. 
Integrating 

-eA  = 

4 


(\  being  a  co  istant  of  integration. 
Intcgr.'i'^.ng  again, 

W 
-  l^h  =  — -'t-'  +  c,x (3) 

There  is  no  constant  duo  to  the  last  integration,  as  x  and  y 
vanisli  together. 
Ihtzvct'H  C  and  A, 


Integrating  twice, 


dv 

-  El-^y- 

dx 


W 


{l-x)'-\-c. 


and 


W 


-  Ely  =  — (/  -  xy  +  c,x  + 


(4) 

(5) 
(6) 


c,,  c,  being  constants  of  integration. 

The  tangent  at  C  is  no  longer  horizontal,  but  makes  a  dcfi- 

dy 
nite  ancrle  f^  with  the  horizon,  so  that  -,-  is  now  tan  0  when 
"^  dx 

•      I  dy 

X  ■=.  -.    Also,  the  values  of  -7-  and  I'  at  6,  viz.,  tan  d  and  d,  as 
2  dx  ' 

given  by  eqs.  (2)  and  (3),  must  be  identical  with  those  given  by 

eqs.  (5)  and  (6),  while  the  value  jj',  at  A,  as  given  by  eq.  (6) 

when  x  ■=.  l,'\?,   equal  to  y^ .     Therefore 


W 
W 


A'  ^c,  =  -EI  tan  e  =  -  ^P  +  c, , 
16       '      '  16       '     ^ 


and 


/  IV  I 

go  2  96  2 

-  Ely,  =  cj -jr  c,. 


I  ill 


i 


h 


448 

Hence, 


THEORY  OF  STRUCTURES. 


V         W 

c  —  ~  EI—  —  --/'   c 


/     16 

fully  defining  both  halves  of  the  neutral  axis. 


V        W  W 


dy 


Again,  in  Ex.  6  it   is  no  longer  true  that  ->-  =  o  when 

ax 

I 
X  ■=.—,  but  the  conditions  of  integration  are  j  =  o  when  x  =  0, 

dy 
and  y  =  j,  when  x  =  I.     These,   together  with  ^^  =  o  when 

X  =^  o,  are  also  the  conditions  in  Ex.  7.     Other  cases  may  be 
similarly  treated. 

5.  To  Discuss  the  Form  assumed  by  the  Neutral  Axis 
of  a  Girder  OA  which  rests  upon  Supports  at  o  and  A, 
and  carries  a  Weight  P  at  a  Point  JB,  distant  r  from  o. 

Let  OBA  be  the  neutral  axis  of  the  deflected  girder. 

r  ,  The  reactions  at  O  and  A  arc 

.,,—'         ^       I  —  r  r 

P — -, —  and  P-.,  respectively. 

Y         '  Let    BC,    the   deflection   at  C 

F'G.  3"-  =  d. 

Let  a  be  the  slope  of  the  neutral  axis  at  B. 
The  portions  OB,  BA   must  be  treated  separately,  as  the 
weight  at  B  causes  discontinuity  in  the  equation  of  moments. 
First,  at  any  point  {x,  y)  of  OB, 


---iB 


-EI 


dy 

dx' 


-X, 


(I) 


Integrating, 


^4=^^ 


rx 


+  ^„ 


r,  being  a  constant  of  integration. 


NEUTRAL  AXIS  OF  A   LOADED  BEAM. 


449 


But  -^  =  tan  Of  when  x  =  r,  and  therefore 
ax 

I  -rr' 
—  EI  tan  a  =  P  — J-  -  +  <^i  • 


Hence, 


hitcgrating, 


/  —  r  fx"      /•'    \ 
-  E/{y  -  X  tan  a)  =  P  —^  [-^  -  -xj. 


(2) 


(3) 


There  is  no  constant  of  integration,  as  x  and  j/  vanish  to- 
gether. 

Also,  f  =  d  when  x  =  r. 


EI{d  -  r  tan  a)  =  -  P 


I  -rr' 


I     l      '     - 


(4) 


In  the  same  manner,  if  A  is  taken  as  the  origin,  and  AB 
treated  as  above,  equations  similar  to  (i),  (2),  (3),  and  (4) 
will  be  obtained,  and  may  be  at  once  written  down  by  sub- 

v-.^  /  —  r 

stituting  in  these  equations  n  —  a  for  oi,  P-.  for  P — -. — ,  I—  r 

for  r,  and  r  for  I  —  r. 

Thus,  the  equation  corresponding  to  (4)  is 


-EI\d-{l  -  r)  tan(;r-a)|  =  -  P 


I       3 


Subtracting  (5)  from  (4), 


P 

Ell  tan  a  =  -r{l-  r){l 

0 


and  from  (4), 


£/d  =  - 


Pr\l-ry 
3         /       • 


(5) 


—  2r) ;   .    .    .    .    (6) 


(7) 


■'.^'  ',1' 

11 

■ii 

;    1 
j    i 

■        ) 

:    I 

: 

i^  t  ii 


i,  a  h 


'i  f 


Jl:- 


450  THEORY  OF  STRUCTURES. 

Thus,  eqs.  (2)  and  (3)  become 


and 


^dy       PI  -r  ^       Pr  ,, 


^^y     =6-7-^^°-67(^-'')(2^-''K-    •    (9) 


the  latter  being  the  equation  to  the  portion  OB  of  the  neutral 
axis,  and  the  former  giving  its  slope  at  any  point. 
Next,  at  any  point  {x,y)  of  BA, 


ipy          I  —  r 
-EI-:ii,=iP—y-x-P(x-r) (10) 


dx' 


I 


Integrating, 


ax  /     2      2^         J  ^  t> 


c^  being  a  constant  of  integration. 

But  -7-  =  tan  a  when  ;f  =  r. 
dx 


I  -rr" 


Pr 


I      2 


-\-c^=  —  £/  tan  a  = li/ —  ^)i^  —  2r), 

3  * 


and 


Hence, 


Pl-r 
6 


<:.=  .-^— -r(2/-r). 


^4  =  fV-'-^(^-'')'-?7(^-''X^^-^)-^") 


Integrating, 

^,  being  a  constant  of  integration. 
But  y  —  d  when  x  =  r. 


4> 


NEUTRAL   AXIS  OF  A   LOADED  BEAM. 


45  X 


P  I  —  r         P  r 
.-.  g- -^  r' - -g- -j(/ -  r)(2/ -  r>  +  r, 


=  -  Eld  =  - 


Pr\l-ry 


and  c^  =  o.     Hence, 


P  I  —  r         P  P  r 

-EIy  =  -^  —j-x'  -  -Ax  -rf  -p-  -j{l  -  r){2l  -  r)x,      ( 1 2) 


/ 


6  r 


which  is  the  equation  to  the  portion  BA  of  the  neutral  axis, 
eq.  (11)  giving  its  slope  at  any  point. 

In  the  figure  r  <  — ,  and   the   maximum  deflection   of  the 

girder  will  evidently  lie  between  B  and  ^,  at  a  point  given  by 

dy 
putting  -j~=^  o  in  eq.  (12),  which  easily  reduces  to 


—  2lx-\- 


2/'  + 


=  0, 


and  therefore 


--4 


r  -r' 


3 


IS  the  abscissa  of  the  most  deflected  point.     The  corresponding 
deflection  is  found  by  substituting  this  value  of  x  in  eq.  (12). 

If  r  >  — ,  the  maximum  deflection  lies  between  O  and  B,  at 
2  ' 

dy 
a  point  determined  by  putting  -3-  =  o  in  eq.  (8),  which  then 

easily  reduces  to 

r(2/  -  r) 


0  =  X' 


from  which 


.  =  ^fc> 


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452  ••  THEORY  OF  STRUCTURES. 

Substituting  this  value  of  x  in  eq.  (9), 

J  n     '           P  l-r(r{2l-  r)\l 
the  maximum  deflection  =  Tp'T~l~  \ "/  " 

Example. — P  =  15,000  lbs.,  /  =  100  ft.,  r  =  90  ft. 
The  distance  of  most  deflected  point  from  O 


sj'^- 


_./?oX^=57^ft., 


and  the  maximum  deflection 


15000   10  ^90  X  I  io\* 


X 


10  ^90  X I  ipy  _ 
100  \    3    /  ~ 


500000,  ,, 


6.  To  Discuss  the  Form  of  the  Neutral  Axis  of  a  Girder 
OA  which  rests  upon  Supports  at  O  and  A  and  carries 
several  Weights  i*, ,  P, ,  P, , .  .  . ,  at  points  i,  2,  3, .  . . ,  of 
which  the  Distances  from  O  are  r, ,  r, ,  r, , . . . ,  respectively. 


:^X 


X 


Fig,  333, 

It  may  be  assumed  that  the  total  effect  of  all  the  weights 
is  the  sum  of  the  effects  of  the  separate  weights,  and  thus  each 
may  be  treated  independently,  as  in  the  preceding  article. 

Let  or, ,  or, ,  or, , ...  be  the  slopes  at  the  points  I,  3,  3, . . . 
of  the  neutral  axis. 

Considering  P^ ,  the  equation  to  0\  is  l  :.'. 

-  Ely  =  -y  ^^x^  ~  ^  7'(/  -  r  ){2l-r,)x; 


NEUTRAL  AXIS  OF  A   GIRDER   OA. 


4S31 


and  to  lA, 


^  Ely  =  §  ^^V  -  ^{x  -  r.)'  -  §  "jil  -  r.)(2/-  r,)x. 


Considering  P^ ,  the  equation  to  O2  is 

-  Efy  =  §  ^V  -  §  ^(/  -  r,)(.2/  -  r,)A: ; 
and  to  2A, 

-  £/;'  =  §  ^-=^V  -  ^(x  -  r.)'  -  §  "jil  -  r,)(2/  -  0;r  ; 

and  so  on  for  P, ,  P, ,  etc. 

The  total  deflection  V  at  any  point  {x,  Y)  is  the  sum  of  the 
deflections  due  to  the  several  loads. 

Take,  e.g.,  a  point  between  3  and  4,  and  let  d?, ,  </, ,  rf, , .  .  . 
be  the  deflections  of  this  point,  due  to  P,  ,/*,,/*,,...  ,  respec- 
tively.   Then 

-Eld,  =  §^-^-«^'  -  -/f!  -  r.Xa/-  r,)x  -  -^{x  -  r.)'; 

-Eld,  =  §  ^-^p;.'  -  §  ^X/  -  r,)(2/  -  r>  -  5(^  -  r.)' ; 

-£M  =  §  ^-^V  -  §  ^^'(Z  -  r.)(2/  -  r>  -  ^\x  -  r,y ; 

-£M  =  ^*  ^V  -  §  ^X/  -  0(2/  -  r> ; 

and  so  on.     Hence, 

-  EIY=  -  EI{d,  +</,  +  ...) 


■;j..   f  '        ' 

;  1 

"'■Ml    '" 


'^ii 

'  -.-Wi 

m 


454 


THEORY  OF  STRUCTURES. 


i 


1 


Again,  the  position  of  the  most  deflected  point  is  found  by 
making  -^f-  =  o  in  the  equation  to  that  portion  of  the  neutral 

axis  between  two  of  the  weights  in  which  the  said  point  lies. 
The  result  is  a  quadratic  equation,  and  the  value  of  x  derived 
therefrom  may  be  substituted  in  eq.  (A),  which  then  gives  the 
maximum  deflection. 

Example. — A  girder  of  loo  ft.  span  supports  two  weights 
of  20,000  lbs.  and  30,000  lbs.  at  points  distant  respectively  20  ft. 
and  60  ft.  from  one  end. 

The  most  deflected  point  must  evidently  lie  between  the 
two  weights,  and  the  equation  to  the  corresponding  portion  of 
the  neutral  axis  is 

^^,^       X*  ,                „     ,                      .       20000, 
—  EIY  —  g— (20000  X  80  +  30000  X  40) 6~"(^  ~  20) 

—  g--(2oooo  X20X80  X  180+ 30000x60x40 X  140) 

14000   ,        lOOOO,  ,.  . 

=  -^ X {x  —  20)  —  26400000:^. 


F  is  a  maximum  when 


dY 


■j-=.Q—  i4000;r'  —  ioooo(;jr  —  20)*  —  26400000, 


I": 


or 


or 


.r*  -f  lOO;jr  —  7600  =  O, 


.*•  =  50.497  ft. 

EI  d*y 

Remark. — Instead  of  assuming  -^  —  ±^EI~^  ,  it  would  bt 

more  accurate  to  take  -^  =  ±  EI  cos  6  —  (Art.  2),  and  the 

first  integration  would  make  the  left-hand  side  of  the  slope 
equation  ±  EI  sin  6  instead  of  ±  EI  tan  0. 


i.r!, 


MOMENT  OF  INERTIA  VARIABLE. 


455 


7.  Moment  of  Inertia  variable. — In  the  preceding  exam- 
ples the  moment  of  inertia  /  has  been  assumed  to  be  constant. 
From  the  general  equations, 

ax        c 

c  being  proportional  to  the  depth  of  the  girder  at  a  transverse 
section  distant  x  from  the  origin. 

Hence,  for  beams  of  uniform  strength,  the  value  of  c  in 
terms  of  x  may  be  substituted  in  the  last  equation,  which  may 
tlien  be  integrated. 

Again,  let  Fig.  323  represent  a  cantilever  of  length  /,  spe- 
cific weight   If,   circular  section, 
and  with  a  parabolic  profile,  the 
vertex  of  the  parabola  being  at  A. 

Let  2b  be  the  depth  of  the 
cantilever  at  the  fixed  end. 

Let  the  cantilever  also  carry  a 
uniformly  distributed  load  of  in- 
tensity/. Fio.  323. 

Consider  a  transverse  section  of  radius  ^  at  a  distance  x 
from  the  fixed  end. 

Let  X,  y  be  the  co-ordinates  of  the  neutral  axis  at  the  same 
section.     Then 


EI 


j{l-xy+^-{l-xr  =  E  — 


n£^(Py 
4    dx*' 


b\ 


But2'  =  ^-(/-.»r). 


itb*'  d*v      wit  b*  t) 


or 


nE  b*  d*y  _  w  w  ^'  \  _\    P 


Integrating, 


nEb^dy       wnfl,        x*\       fix 


J 


'■•^i 


ri 


w 


n 


■Set- 
.!■     I 

A  ■  \ 


456 


THEORY  OF  STRUCTURES. 


dy 


There  is  no  constant  of  integration,  as  ;/-  =  o  when  ;r  =  0. 
Integrating  again, 


2) 


There  is  no  constant  of  integration,  as  x  and  y  vanish  to- 
gether. Thus,  equation  (i)  gives  the  slope  at  any  point,  and 
equation  (2)  defines  the  neutral  axis. 

•    The  slope  at  the  free  end  {x  =  I)  =  -=^{ — | — tj-J. 

The  deflection  "        "  «        =  T^r.!-"^  +  -4-). 

£d\g      '   no  J 

8.  Springs  Fixed  at  One  End  and  Loaded  at  the  other 
with  a  Weight  w. 

Data. — Length  =  /;  breadth  =  6,  and  depth  =  d  at  fixed 
support ;  V=  volume  of  spring;  /=  maximum  coefficient  of 
strength  ;  J  =  maximum  deflection. 

Case  a.  Simple  rectangular  spring. 

By  Ex.  I,  Art.  39, 


iW 


i 


ll 


since 


Fig.  324. 

Also, 

WA 
Hence, 


I  1^/'  _  2  //* 
-  lEI-~i~Ed' 


Wl  _M _2f  _  12WI 
~  I  ~  I  ~  d  ~   bd*  ' 


(I) 


bd'f  2fll_\  £bdl  _  £V 
61   '  zEd~'^    E     ~  gE' 


V  =  9 


WAE 


r 

WA       PV 

The  work  done  = =  ~t. 

2         i8£ 


•        •        • 


•         •         •        • 


(2) 

(3) 


\\ 


SP  KINGS.  457 

Case  b.  Spring  of  constant  depth  but  triangular  in  plan. 
Let  bx  be  the  breadth  at  a  distance  x  from  the  fixed  end. 
Then 

b^      I  —  X  'A . 

o  I  ^ 

and  /  at  the  same  point  *W 

'''dx'-Er      ^^~  Ebd*' 

Integrating  twice, 

dy  _  \2Wl 
dx  ~  Ebd'"^' 
and 

_ewi  , 
•■•  ^  -  £bd^  -~Ed ^^^ 

Also, 

^^  ~    61  Ed  ~   6E    ~  3£'     ■ 
Hence, 

...y^^jp..    ........    (5) 

wj   r  V 

The  work  done  =  — ^  =  -^-^ (6) 

N.B. — The  results  i  to  6  are  the  same  if  the  springs  are 
compound  ;  i.e.,  if  the  rectangular  spring  is  composed  of  n 
simple  rectangular  springs  laid  one  above  the  other,  and  if  the 
triangular  spring  is  composed  of  n  triangular  springs  laid  one 
above  the  other. 


>  ■ 


I  *■ 


iUf 


•*-..  t 


^1' 

I'  i 

i     i 

!■'  ^ 

i.  i 


li 


'■    .- 

L 

458  THEORY  OF  STRUCTURES. 

Case  c.  Spring  of  constant  width  but  parabolic  in  elevation. 
Let  d,  be  the  depth  at  a  distance  x  from  the  fixed  end. 

Then 


IW 

r 


Pic.  396. 


{d.\_l-x 
\d]-     I     ' 

and  /at  the  same  point  = 
'       bd:_  _  bjT  il  -  Ar\t 

~2    ~     I2A      /      /• 


Integrating  twice, 


y  = 


12W  I* 

~'Erbd' 


\^{l-x)^-2l\l-x)^\l% 


and  hence 


£    bd'~  lEd' 


(7) 


Also, 


rF-^  = 


^  ^JL^lLMl=-f-W. 
61  '  lEd     <)  E  lE    ' 


.:  V  = 


The  work  done 


WA  _  i/'V 
2    ~6  E  ' 


9.  Girder  Encastrfe  at  the  Ends.— The  girder  BCDEFG 
rests  upon  supports  at  the  ends,  is  held  in  position  by  blocks 
forced  between  the  ends  and  the  abutments,  and  carries  a  uni- 
formly distributed  load  of  intensity  w. 


GIRDER  ENCASTR£  AT   THE  ENDS. 


AS9 


It  is  required  to  determine  the  pressure  that  must  be  devel- 
oped between  the  blocks  and  the  girder  so  that  the  straight 
portion  between  vertical  sections   at  points  O  and  A  of  the 


R 

I 
D  1 


"^^^m 


-t— i 


i  01 


^^pTa 


1 1  ^^ — 


■  t' :  i   i    i 

!.    ■,■.■1  )i 
■■\\   \:,t    I.  . 


Fig.  327. 

neutral  axis  may  be  in  the  same  condition  as  if  the  girder  were 
Jixcd  at  these  sections. 

Let  /  be  the  length  of  OA. 

Let  R  be  the  reaction  at  the  surface  BC,  and  r  its  distance 
from  O. 

Let  H  be  the  reaction  between  the  block  and  the  end  CD, 
and  k  its  distance  from  O. 

Let  P  be  the  weight  of  the  segment  on  the  left  of  the  ver- 
tical section  O,  and/  its  distance  from  0. 

Then  for  the  equilibrium  of  the  segment  on  the  left  of  the 
section  at  O, 

wl                                                               tvl^ 
R4---P=o,     and     Rr  -  Pp  -  Hh •  =0. 

'2  ^  12 

.■.R  =  P--. 


and 


H  = 


I         wis  _        tvl* 


h 


=  the  required  pressure. 


#1 


Again,  take  O  as  the  origin,  OA  as  the  axis  of  x,  and  a 
vertical  through  O  as  the  axis  of  y. 

At  any  point  (;r,  y)  of  the  neutral  axis, 


d^y        wl  WX*        wl* 

EI~,  =  ~x-  —  -  — 
dx^        2  2  12 


(See  Ex.  8.) 


I'll 


I 


1 


■  ! 


!iti 


n 


u 


'•'!    1 


.'  ) 


460 


THEORY  OF  STRUCTURES. 


10.    On   the  Work   done   in   bending:   a   Beam. — Let 

^  A'B'C'D'  be  an  originally  rectangular  ele- 
^'  ment  of  a  beam  strained  under  the  action  of 
Q'   external  forces. 

Let  the  surfaces  A' D' ,  B'C  meet  in  0\ 
O  is  the  centre  of  curvature  of  the  arc  P'Q  of 
the  neutral  axis. 

Let  OF  =  R=  OQ'. 

Let  the  length  of  the  arc  P'Q'  =  dx. 

Consider  any  elementary  fibre /»y,  of  length 
dx',  of  sectional  area  a,  and  distant  y  from  tlie 
neutral  axis. 

Let  /  be  the  stress  \np'q'. 

The  work  done  in  stretching/'^' 


=z  ^~-[dx'  -  dx^. 


But 


dx  ~  P'Q' 


R-\-y          ,             ^   dx' —  dx       ^    y 
and    t  —  Ea ; =  Ea  j,. 


R 


dx 
\E 


The  work  done  in  stretching/'^'  =  —  -w^dxay'', 
and  the  work  done  in  deforming  the  prism  A'B'C'D' 
=  ^[-^^^dxay')  =  '-^,dx2iaf)  ^\^.  dx. 


Hence,  the  total  work  between  two  sections  of  abscissx 


A'  \ET  EI  C* 


I       M 
But  -^  —  -py;  therefore  the  work  between  the  given  limits 


EI  /*"' 


'^^%)    =2EI 


m^m 


TRANSVERSE    VIBRATIONS. 


461 


This  expression  is  necessarily  equal  to  the  work  of  the  ex- 
ternal forces  between  the  same  limits,  and  is  also  the  semi  vis- 
viva  acquired  by  the  beam  in  changing  from  its  natural  state 
of  equilibrium. 

Ccr. — If  the  proof  load  P  is  concentrated  at  one  point  of  a 

P 

beam,  and  if  d  is  the  proof  Jeflection,  the  resilience  =  —d. 

If  a  proof  load  of  intensity  zv  is  uniformly  distributed  over 
the  beam,  and  if  y  is  the  deflection  at  any  point,  the  resili- 
ence =  ~  /  wydx,  the  integration  extending  throughout   the 

whole  length  of  the  beam. 

The  case  of  the  single  weight,  however,  is  the  most  useful 
in  practice. 

II.  On  the  Transverse  Vibrations  of  a  Beam  resting 
upon  Two  Supports  in  the  same  Horizontal  Plane. 

It  is  assumed — 

(a)  That  the  beam  is  homogeneous  and  of  uniform  sectional 
area. 

{b)  That  the  axis  {neutral)  remains  unaltered  in  length. 

{c)  That  the  vibrations  are  small. 

[d)  That  the  particles  of  the  beam  vibrate  in  the  vertical 
planes  in  which  they  are  primarily  situated.  In  reality,  these 
particles  have  a  slight  angular  motion  about  the  horizontal  axis 
throup[h  the  centre  of  gravity  of  the  section,  but  for  the  sake 
of  simplicity  the  effect  of  this  motion  is  disregarded. 


dx 


A-X 


Ci;C' 
i  s+as 
vodx 


fY 


Fig.  329. 

Let  OA  be  the  beam. 

Take  O  as  the  origin,  the  neutral  line  OA  as  the  axis  of  x, 
and  the  vertical  C>Fas  the  axis  o[  y. 

Consider  an  element  of  the  beam,  bounded  by  the  vertical 


SbHI^H 

:eay 

^^^^^H 

KflP 

tf  1 

w 

[ 
1* 

1 

L 

t 

X   ■ 

■j  ■ 

,t-  ; 

■  n- 

^^ 

"■El' I? 


■■^ 


FUL. 


1  ;■ 


I,      I! 


409 


THEORY  OF  STRUCTURES. 


planes  BC,  B'C\  of  which  the  abscissae  are  x  and  x  -\-  dx, 
respectively. 

Let  w  be  the  intensity  of  the  load  per  unit  of  length ;  hence 
wdx  is  the  load  upon  the  given  element,  and  acts  vertically 
through  its  centre. 

Let  S  be  the  shearing  force  at  .5 ;  S  -{•  dS  the  shearing 
force  at  B'. 

Let  M  be  the  bending  moment  at  B\  M-\-dM  the  bend- 
ing moment  at  B'. 

Also,  the  resistance  of  the  element  to  acceleration  =  -  ~  -. 
Hence,  at  any  time  t. 


w      d^v 

-dx  -^  +  5  -  (5+  ^5)  -  wdx  =  o, 

O 


or 


dy      g  dS 


df 


w  dx 


-g=o. 


(I) 


Again,  taking  moments  about  the  middle  point  of  BB'  or 
CC, 


M.il 


4' :;!' 


M-{M  +  dM)  +  S~  +  {S-{-dS)~  =  o, 


or 


^dy 


dM 
'dx 


=  S. 


But  iW  =  —  £/-T=i .     Therefore 
dx 


2 


•  «  •  •  < 


(2) 


dx' 


dS 


dy 


•^=-^^^'  ^"^  di=-^^d-^^ 


Hence,  from  (i). 


5+^£/~^-^=o. (3) 

dt     '   IV       dx*      ^ 


CON^TINUOUS  GTRDERS. 


463 


This  equation  docs  not  admit  of  a  finite  integration,  but 
may  be  integrated  in  the  form  of  a  partial  differential  equation. 

12.  Continuous  Girders. — When  a  girder  overhangs  its 
bearings,  or  is  supported  at  more  than  two  points,  it  assumes  a 
wavy  form  and  is  said  to  be  continuous.  The  convex  portions 
are  in  the  same  condition  as  a  loaded  girder  resting  upon  a 
single  support,  the  upper  layers  of  the  girder  being  extended 
and  the  lower  compressed.  The  concave  portions  are  in  the 
same  condition  as  a  loaded  girder  supported  at  two  points,  the 
upper  layers  being  compressed  and  the  lower  extended.  At 
certain  points,  called  points  of  contrary  flexure,  ox  points  ?f  in- 
flexion, the  curvature  changes  sign  and  the  flange  stresses  ai'- 
necessarily  zero.  Hence,  apart  from  other  practical  considera- 
tions, the  flanges  might  be  wholly  severed  at  these  point  vith- 
oiit  endangering  the  stability  of  the  girder. 

13.  The  Theoreii  of  Three  Moments. — It  is  required  tw 
determine  a  relation  between  the  bending  tuoments  at  any  •  >t, 
consecjiiix'C  ^  Jnts  of  support  of  a  loaded  continuous  girder  of 
several  spans. 


Rr-l 

A 


Rr 

iX 


V 


Rr+1 


Fig 


330. 


Let  O,  X,  V  be  the  (r  —  i)th,  rth,  and  (r+  i)th  supports, 

respectively. 

Let   OX=l^,XV-lr^r' 

Case  A,  Let  w^  be  the  load  per  unit  of  length  on  OX, 
tcv+i  the  load  per  unit  of  length  on  XV. 

Let  Rr-i,  Rr,  Rr  +  i  bc  the  reactions  at  0,X,  V,  respectively. 

Let  Mr-,,  Mr,  Mr^i  be  the  bending  moments  at  O,  X,  V, 
respectively. 

Let  a  be  the  angle  which  the  tangent  to  the  girder  at  X 
makes  with  0  V. 

Consider  the  segment  OX,  and  refer  it  to  the  rectangular 
axes  Ox,  Oy. 


\ 

i 


M 


;{' 


u 


Mil 


464  THEORY  OF  STRUCTURES. 

The  equation  of  moments  at  any  point  {x,  y)  is 

-  EI-^.  =  Rr-,X  -  w,  y  +  M,.,  =AI.        .      .      (I) 

At  X,  X  =  I, .  and  M  —  M, . 


/' 


.       ^) 


c; 


imilarly,  the  segment  A^F  gives 

^,.^,/,  +  ,-7C,+  ,-^+J/,  +  ,  =il/,.  .       .      (3) 

Combining  (2)  and  (3), 


=  MXlr  +  ^r  +  ■) (4) 


Integrating  (1), 


-T^y 


EI 


d. 


V  X  X 

,-   =  R,.,  V  ~   ^-  "0    +  ^r-i^  +  <^,         .        .       (5) 


c  being  a  constant  of  integration. 

dv 
When  X  —  I, ,  -',--  —  tan  a. 
dx 


Ir  ly 

.'.  —  EI  tan  ix  —  AV_,  --  —  iv,.  —  +  ^r-Jr  +  ^>,      •     •    (^) 

Integrating  (5), 

X*  V*  x^ 

-  Ely  =  Rr_,  -^  -Wr '--  4-  J/,.,  --  +  ex.      .    (7» 

There  is  no  constant  of  integration,  as  x  and  y  vanish  together. 
Also,  >'  =  o  when  x  =  l, . 

■■■o  =  xJ^-w/£  +  mJ-^  +  J„ 


or 


.=:-/?,., ^  +  «V^-i^/.-i'.     ...       (8) 


-t 

Ip    '^ 

ii 

; 

1^1 

iL.a 

i  ,L 

^^F"W^P 


™"l  mT*  « 


THE    TIIEOKEM  OF    THREE   MOMENTS. 


Substituting  this  value  of  c  in  cq.  (6), 

-  EI  tan  a  =  /?,.,  -^  -  w,  "  +  M,.,  ~. 

Similarly,  the  segment  A'f^  gives 


465 


(9) 


lif  tan  (;r  -  ..)  -  R,  ,  .  ^^  -  w,. ,  /  '^^  -f  ^r!'-^ . 


do) 


Adtling  cqs.  (9)  and  (lo),  transposing,  and  simplifying, 

=   IWJ;  +  |«',  +  A   ,    .    -    IMr.Jr   -    IMJ,,.,.        (I  I) 

Finally,  combining  cqs.  (4)  and  (11), 

M,_J,.  +  2ArXlr  +  Ir,)  +  ^/m/h.  =    -    aw:  -f  7t/.  +  ,  /V,).  (12) 

If  the  girder  is  sujjported  at «  points,  there  are  n  —  2  equa- 
tions connecting  the  corresponding  bending  moments,  and  two 
additional  equations  result  from  the  conditions  of  su])port  at 
the  enils.     For  example,  if  the  ends  merely  rest  upon  the  sup- 

(iy 
ports  !/■,  '—  O  and  M„  :=  o  ;  if  an  end  is  fixc^d,  -,-    ::=  o  at   that 

point. 

The   point  of  maximum  bending  moment,  the  points  of 

inflexion,  and  the  point  of  maximum  deflection  in  any  span  are 

,.      dM  - .  <iy 

found  by  making  -.— -  —  o,  AI  =  o,  and         —  o,  respectively. 

Thus,  for  the  span  OX, 


dM 
dx 


o  =  Ry.^  —  WrX\ 


R  I  R* 

:.  X  =  ~^',    and  maximum  IJ.M.  = ^^  4-  J/_., ; 


M=.0  =  Rr-iX 

a  quadratic  giving  x ; 


w,---f  iT/,_., 


il 


■  M 


it  r 

ii 
Ml' 


!  t 


ills 

I. 
I 


Uh- 


: 


I 


?;,  ' 


I  ?''• 


•?, 


K 


466 


THEORY  OF  STRUCTURES, 
dy  X*  x^ 


.1  cubic  from  which  1  may  be  found  by  trial.  The  maximum 
(irtlcction  is  obtained  by  substituting  the  value  of  x  in  eq.  (7) ; 
(•  being  given  Ijy  cq.  (8). 

Case  B.  Let  the  loads  upon  OX,  XV,  respectively,  consist  o! 

a  number  of  weights  /',  ,J\,1\ distant  /.  ,p,,/y. fmni 

O,  and  C^, ,  C\ ,  (?3 ,  .  .  .  .  distant  </, ,  q.,  ,</,,...  from  K.  Refer 
the  neutral  axis  OAA"  to  the  rectangular  axes  Ox,  Oy. 

It  may  be  assumed  that  the  total  effect  of  all  the  weights 
is  the  algebraic  sum  of  the  effects  of  the  weights  taken  sep.i- 
ratcly. 

Consider  the  effect  of  P,  at  A. 

The  equation  of  moments  at  any  point  {x,  y)  of  the  neutral 
axis  between  O  cxud  A  is 


EI  -/,  =  R,_,x  -f  M,. 

dx  »•  ■       I 


(I) 


Integrating, 


E-i%^Rr-f-^M,.,X^C,,         ...       (2) 


Cy  being  a  constant  of  integration. 
Integrating  again, 


x^  X* 

-  Ely  =  /e,_.-g  +  Mr.,~-  4-  <^.^. 


(3) 


There  is  no  constant  of  integration,  as-r  and  _;/ vanish  together. 
The  equation  of  moments  at  any  point  {x,y)  between  A  and 
X\s 

-El'^j^,  =  K.,x-Plx-p,)  +  Mr.,.    .    (4) 


liitc-jratiiii^  ■i^aiii, 


Ely  =  AV. .:-  -  -^{x  -  p,r  +  M^J-  +  c,x  +  c, .       (6) 


(fy 
Now,  at  the  point  A,  the  values  of  -,     and  j,  as  given  by 

cqs.  (5)  and  (6),  arc  identical  with  those  given  by  eqs.  (2)  and 
(3) ;  also,  in  equation  (6),  j  =  o  when  x  —  4. 
Hence, 


.    A' 


K 


AV_,^  +  y?/._,/>,  +  <^.      =  ^r-,-r-  +  ^/.-.  A  +  ^, , 


^■■PWfWpf 


Tin:    THEOREM   OT   THREE  MOMENTS. 


467 


Integrating, 


dy  x''      P 


Kjf  +  ^/.-.^  +  c,p,  =  R.J^-  +  M^Jf  -\-c,p,+c,, 


and 


so  that 


and 


o  =  AV_,|-  -  ^'(4  -  /.)'  +  ^/.-.-;-  +  cj,  +  ^. ; 


f,  =  0, 


C,  =  f. 


/'     p 


/. 


6+'6fy'-^^'>^-^^-i"  '  '  (7) 


Let  a  be  the  slope  at  A';  then,  by  eqs.  (5)  and  (7), 

-  £/tan  a  =  i?,./^  -  ^^(/.  _^,)X24 +/.)  +  .»/.-. ^-^   (8) 


Similarly,  the  segment  XV  gives 

-  £/  tan  {tt  -  a)  =  kJ-^  +  i1/.+,^". 


(9) 


il  ' 


I  Al 


! 


r     f    i 


468  THEORY  OF  STRUCTURES. 

Adding  eqs.  (8)  and  (9),  and  transposing, 


K_,K^  +  Rr^rr^.  =  -%lr  -  P.)\2lr  +/.) 


^1 


-|J/,.,/,-|J/,+,/,+,.        (10, 

Again,  taking  moments  about  X, 

whence 

R^  _,/,»  +  /?.+./%+,  =  M^{1,  +  /,^,)  -  M^_J^ 

-Jf,+,/^,  +  P./.(4-A).     (12) 

and  finally,  by  eqs.  (lo)  and  (12), 

JC/. +  2J/.(/. +  /.+.)  + ^/.+/.+.  =  -Z'. 7(4' -A')-     (13) 

The  effect  of  each  weight  may  be  discussed  in  the  same 
manner,  and  hence  the  relation  between  i1/,._, ,  J/,.,  and  i^/^^, 
may  be  expressed  in  the  form 


M, 


Ppr 


;_  /,  +  2Mllr  +  4+:)  +  i^/.4..4+:  =  —  ^-,-(/.'  -  P  , 

'■r 


-:^^(/%+,-^').     (14) 

Cor.  I.  The  relation  between  M^_i,  M^,  M^^^  tor  a  uni- 
formly distributed  load  may  be  easily  deduced  from  eq.  (14), 
For  example^  let  a  uniformly  distributed  load  of  intensity:;' 
cover  a  length  2a  {<l^)  of  the  span  OX,  and  let  z  be  the  distance 
of  its  centre  from  0.     Then 

which  reduces  to when  ^  =  a  =  — . 

4  2 


THE    THEOREM  OF   THREE  MOMENTS. 


469 


Cor.   2.   Considering  the   rth  span  and   taking  moments 
about  the  rth  support, 

.)/  being  the  moment  of  the  load  on  the  span,  and  the  reaction, 
(If  shear, 

Hence,  the  shear  at  the  (r  —  i)th  support  for  the  rth  span 
=  the  reaction  at  the  same  support,  supposing  the  span 

an  independent  girder,  i.e.,  cut  at  its  supports, 
-f-  the  difference  of  the  forces,  or  reactions,  equivalent 
to  the  moments  at  the  supports. 
Again,  let  M^  be  the  moment  of  the  load  on  the  segment 
X  with  respect  to  the  point  {x,  y). 

Hence,  the  total  moment  about  {x,  y) 


=  the  moment  at  the  same  point  supposing  the  span 

an  independent  girder, 
-f-  the  reactions  equivalent  to  the  moments  J/^, ,  My , 
multiplied  respectively  by  the  segments  l^  —  x  and  x. 
In  Fig.  331,  C>A' being  the  rth  span,  let  OBX  be  the  curve 


Fig.  331. 


of  bending  moments,  supposing  OX  an  independent  girder,  i.e., 
cut  at  0  and  X.  On  the  same  scale  as  this  curve  is  drawn, 
take  the  verticals  OE  and  XF  to  represent  Mr-^  and  M^ ,  re- 


■ 

I, 
t 

•Mi 

m 

m 

i 

i 


470 


THEORY  OF   STRUCTURES, 


■I  r 


spectively,  and  join  EF.  The  curve  OBX  corresponds  to  the 
portion  \-j  x  —  MA  of  the  above  equation,  and  the  line  EF 

to  the  remainder,  i.e.,  — ^'1 /,.  —  x\-\-  -fx.     The  actual  bend- 

in^  moment  at  any  point  of  OX  is  represented  by  the  algebraic 
sum  of  the  ordinates  of  the  curve  and  line  at  the  same  point. 
which  will  be  the  intercept  between  them,  since  they  represent 
bending  moments  of  opposite  kinds. 

Let  A  be  the  effective  moment  area,  or  the  algebraic  sum  of 
the  areas  for  the  load  and  for  the  moments  at  O  and  X,  and 
let  X  be  the  horizontal  distance  of  its  centre  of  gravity 
from  O. 

Let  Ar  be  the  area  for  the  load,  i.e.,  the  area  of  the  curve 
OBX,  and  let  s^  be  the  horizontal  distance  of  its  centre  of 
gravity  from  O.    Then 


Ax  =  A,z^  +  mJ^  +  ;  (^.  -  ^.-:)/; 
^       3 


=  ^^,  +  iJ/,..4'  +  iJ/,/A 


This  result  will  be  referred  to  in  a  subsequent  article, 
14.  Applications. — Example    i.     Swing-bridges    of   two 
spans  revolving  about  a  single  support  at  the  pivot  pier. 

This  is  a  case  of  a  girder  of  two  spans,  0X{=  /,),  XV {—  /^i, 
resting  upon  supports  at  O  and  V,  and  continuous  over  a 
pier  at  X. 

The  bending  moments  at  0  and  Fare 
0  X  V    both  ;///. 

^        Let  M  be  the  bending  moment  at  X. 
For  a  uniformly  distributed  load. 


T — ■        T" 

Fig. 33a 


2il/(/. +  /,)  =  - i(«;/.' +  «.,//),    or     J/=- 


I    W,l^  +  li\ll 

8        /.  +  /.      ' 


w,  being  the  intensity  of  the  load  on  OX,  w,  that  on  XV. 


SWING-BRIDGES, 

For  an  arbitrarily  distributed  load. 


471 


2  J/(A +  /,)  =  - yi  - />',    or     M=~\^^^^, 

where       ^  =  ^'^.^(/.' - /)     and     B  =  ^^!^ {l^  -  q'). 

Let  K^,  A\,  y?.,  be  the  reactions  at  O,  X,  V,  respectively. 
For  a  2iniforiuly  distributed  load, 


^.  =  -:^+-,  - 


8/AA  +  /,) 


For  an  arbitrarily  distributed  load, 

J.  __^P{^.-P)  I  M  _     P{l,-p)      I   A±B_^ 

'"  A     "'^/;"  /,  2 /,(/,  +  /,)• 

If  7^,  =  o,  or  if  P  and  hence  A  =  o,  then  A,  is  negative. 

So  if  zy,  =  o,  or  if  Q  and  ^ence  i)  =  o,  then  A'^  is  nega- 
tive. 

Hence,  if  either  of  the  spans  is  unloaded,  the  reaction  at 
the  abutment  end  of  the  unloaded  span  is  negative  and  that 
end  is  subjected  to  a  liammcriug  action.  This  evil  may  be 
obviated  : 

(rt)  By  loading  the  spans  sufficiently  to  make  A,  and  R^ 
zero  or  positive. 

This  result  is  attained  for  A, 

\{izvj:'-\-A'ii;Kk>w,i:,  or  if  :s^^^^)>i^±^^, 

and  for  7?, 
if  4«'//,'  +  iw^l:  >  ^A',   or  if   :s^l^^  >  1  ^^^_"^^-^. 


n 


i;  I'  ' 


'ii:t 


n 


472 


THEORY  OF  STRUCTURES. 


{b)  By  using  a  latching  apparatus  to  keep  the  ends  from 
rising. 

(<■)  Ry  employing  suitable  machinery  to  exert  an  upward 
pressure,  at  least  equal  to  the  corresponding  negative  reaction 
upon  each  end,  which  is  thus  wholly  prevented  from  leaving  its 
seat. 

Cor.  I.  When  the  load  is  uniformly  distributed,  the  di>. 
tance  x  of  the  point  of  inflection  in  OX  from  O  is  given  by 


M 


^♦^'l-^  ,       ,  r  2/?, 

O  =  K.x ,  and  therefore   x  = 

'2  w, 


Similarly,  the  distance  of  the  point  of  inflection  in  XV  from 

2R 
V=      ' 


Ui 


If  /,  =  /,  =  /,  then 
And  if  w,  =  «',  =  ti',  then 


7w, 


w. 


16 


-V,     K  = 


16 


/. 


M=-  iwP,    R,  =  iwl  =R,,    R,=  2wl -R,-R,^  fzc/. 


2/?,       -,      2R. 
In  the  latter  case =  |/ 


,  and  thus  a  hinge  may 


be  introduced  in  each  span  at*a  distance  from  the  centre  pier 

equal  to  one  fourth  of  the  span,  without  impairing  the  stability 

of  the  girder.     Hence,  also,  the  continuous  girder  of  two  equal 

spans  may  be  considered  as  consisting  of  two  independent 

girders,  each  of  length  |/,  resting  upon  end  supports,  and  of 

/ 
two  cantilevers  each  of  length  — . 

Ex.  2.  Swing-bridges  with  fwo  points  of  support  at  the 

Ri  R«  R»  R4 

h 


Wi 


W^' 

r 


Fig.  333. 

pivot  pier,  as,  e.g.,  when  they  are  carried  upon  rollers  running: 
in  a  circular  path. 


lii_ 


:! 


I'M 


A  P PLICA  TIONS. 


473 


This  is  a  case  of  a  continuous  girder  of  three  spans. 

Let  /,,/,,  /,  be  the  lengths  of  the  spaces,  7f , ,  tc, ,  w,  the 
corresponding  intensities  of  the  loads,  which  are  assumed  to 
be  uniformly  distributed. 

Let  R^y  A',,  R^,  R^  be  the  reactions  at  the  supports; 
,t/,,  M^,  J/j,  M^  the  corresponding  bending  moments.     Then 

Ml  +  2 J///,  +  4)  +  ^./,  =  -  \(zv,i:  +  T.',/,')  ;     .     ( I ) 

i^u + 2iJ/3(/,  +  A)  +  ^y,  -  -  iKC  +  ^^//)-  •  (2) 

Let  the  ends  of  the  girders  rest  upon  the  supports,  and 
assume,  as  is  usually  the  case  in  practice,  that  the  centre  span 
is  unloaded,  i.e.,  that  7f,  =  o.     Then 

M,—Q    and     M^  =  o. 
From  (i)  and  (2), 

2Mll,^i:)^MJ,  =  -\wj: (3) 

J/,/,  +  2yJ/3(/,  +  A)  =  -  M/3' (4) 

-  2wj:{k  +  h)  +  ^^,/oV, 


and 
Hence, 


'^'-4(4//.  +  3C-h4/./3  +  4V,)' 


and 


M^  = 


zvJ:1,  -  2zv,/,V,  +  /,) 


Taking  moments  about  the  second  support, 
RA  =  ^  +  M, 

-  -^e>.(6/.V,  +  6/.V3  +  6/X  +  8/.Vy,)  +  ri'JX 

4(4//,  +  3/,'  +  4//.  +  M) 

Taking  moments  about  the  third  support, 

_  w,{6/X  +  6/.V.  +  6/,V,'  +  8/3V/,)  +  w,/X 

4(4/,/,  +  3/;'  +  M  +  4/y,) 


(5) 
(6) 


(7) 


(8) 


:  i  i 


474 


THEORY  OF  STRUCTURES. 


Thus  /?,  and  R^  are  both  positive  for  all  uniform  distribti 
tions  of  load  over  the  side  spans,  and  no  hammering  action 
can  take  place  at  tiie  ends. 

Again,  if  the  span  on  the  left  is  unloaded,  i.e.,  if  w,  =  o, 
7^/,  is  positive  and  J/,  negative ;  and  if  the  span  on  the  riglit  is 
unloaded,  i.e.,  if  tc,  =  o,  iJ/,  is  negative  and  M,  positive. 

Thus,  at  the  piers,  the  flanges  of  the  gink  r  will  be  sub- 
jected to  stresses  which  are  alternately  tensile  and  compressive, 
and  must  be  designed  accordingly.  The  same  result  is  also 
true  for  arbitrarily  distributed  loails. 

Ex.  3.  The  weights  on  the  wheels  of  a  locomotive  passin<j 
over  a  continuous  girder  of  two  spans,  each  of  50  ft.,  taken  in 
order,  arc  as  follows  :  15,000  lbs.,  24,000  lbs.,  24,000  lbs.,  24,000 
lbs.,  24,000  lbs.  The  tiistances  of  the  wheels,  centre  to  centre, 
taken  in  the  same  order,  are  90  in.,  56  in.,  52  in.,  56  in.  Let 
it  be  required  to  place  the  wheels  in  such  a  position  as  to  give 
the  maximum  bending  moment  at  the  centre  pier. 

The  pier  must  evidently  lie  between  the  third  and  fourth 
wheels. 

Let  X  be  the  distance,  in  inches,  of  the  weight  of  15,000  lbs. 
from  the  nearest  abutment.  The  remaining  two  weights  on 
the  span  are  respectively  ;ir  -f-  90  and  x  -}-  146  in.  from  the 
same  abutment. 

The  two  weights  on  the  other  span  are  142— .r  and  198— .t 
in.,  respectively,  from  the  nearest  abutment. 

Hence,  by  Case  B,  Art.  13,  if  iW  is  the  bending  moment  at 
the  centre  pier, 


-4iI/X  600=  i5^^(6oo'-;r')+^-^(;.r+9o)  1 600'  -  {x  +  90)' 

+  ^-g^(^+H6)16oo'-(.r+i46y( 


24000 
+  "65^  X(i42-4r)|6oo'-(i42 


2-X) 


+  '-^X(i98-;^){6oo'-(i98-.tf 


wm3 


APPLICA  TIONS. 


475 


dM 


Making  -7—  =  o  for  maximum  value  of  M,  and  simplifying, 
I5;ir' 4- 276484:=:  2518848, 


and  therefore 


X  —  87.39  in.  =  7.28  +  ft. 


Tluis,  the  R  M.  at  the  centre  pier  is  ;i  maximum  when  the 
first  wheel  is  7.28  ft.  from  the  nearest  abutment. 

The  maximum  B.  M.  in  inch-pounds  is  obtained  by  substi- 
tuting X  ~  87.39  in.  in  the  above  equation. 

15.  Maximum  Bending  Moments  at  the  Points  of  Sup- 
port of  Continuous  Girders  of  u  equal  Spans. 

Let  the  figure  represent  a  continuous  girder  of  ft  spans,  I. 
2,  3,  .  .  .  «  —  I  being  the  n  —  i  intermediate  supports. 

0123 r—  I     r    r-\- 1 ti—i     n 

*^    7r~'A       A~~A       A       AAA      A      A      A~"r^ 

Case  I.  Assume  all  the  spans  to  be  of  the  same  length  /, 
and  let  w, ,  w, ,  .  .  .  w^., ,  %v„  be  the  intensities  of  loads  uni- 
formly distributed  over  the  ist,  2d,  .  .  .  {ji  —  i)th  and  //th  spans, 
respectively. 

By  the  Theorem  of  Three  Moments, 


4;«,  +  m,  ~  --{w,-\-w^\. 


4 


4 


4 


m,  +  4m, -\r  m,  =  -  -{w,  -f  w,). 

4 


(I) 

(2) 
(3) 
(4) 
(5) 


' 

FT 

'   1 

1 

,  1 

1 

1 

' 

I 


|:r 


476 


THEORY  OF  STRUCTURES. 


w„_,  4-  4w,_. 


-  -•(w„_,-{-7f„).  («-l| 

4 


w„  and  ;;/„  are  both  zero,  as  the  girder  is  supposed  to  be  rest- 
ing upon  the  abutments  at  o  and  n. 

From  tliese  («  —  i)  equations,  the  bending  moments  w, , 
w, ,  .  .  .  ;//„.,  may  be  found  in  terms  of  the  distributed  loads. 

Eliminating  ;«,  from  2  and  3, 

m,  -  1 5'«,  -    4'«4  =-.!(«'«  +  w.)  —  4(«',  +  w.)}.  •     .    {x) 
Eliminating  ;«,  from  4  and  x^ , 

Eliminating  m^  from  5  and  x^, 
ntj  —  2ogm^  —  56W, 

4 
Finally,  by  successively  eliminating  m^,  tn^,  .  . .  m„_2 , 

r 

=  -TJC^^  +  w,)  -  4(^3  +  ^4)  +  I  SK  +  w.)—  .  .. 
4 

the  upper  or  lower  sign  being  taken  for  the  terms  within  the 
brackets  according  as  Jt  is  odd  or  even,  and  the  coefificients 
^w-i  >  ^1.-2  >  ^n-3 .  •  •  •  being  given  by  the  law, 

a„-i  =  4^n-a  —  «»-3  ; 


m 


ATPUCA  TIONS.  477 

«.  =  4^.  —  rt,  =  209  ; 

<».  =  4«, -<^=    15; 
«.  =  4^.  =     4 ; 

a,  =  I. 
Commencing  with  equations  w  —  3  and  «  —  2,  and  proceed- 

iiig  as  before, 


-_^_^ 


an-l^,  +  ^,)  •-     "— 3(^1  +  «^a)  +  an-Sw^  +  W,)  —  .  .  . 


±  1 5(w«_4  +  w«-3)  T  4(«^»-3  +  "^n-^  ±  (^«-»  4-  w„_,)  [  ,     (2) 

the  upper  or  lower  sign  being  taken  for  the  terms  within  the 
brackets  according  as  n  is  odd  or  even. 
Solving  the  two  equations  j  and  j, 

-  (a„_,rt„_3— rt'„_,^„.^  — 3)zy3  +  .  .  .  :f(3««-.  +  <^»-4  — ««-3)^«-a 

and 

±w»-i(«'«-,  —  0  = 

/' ( 

4  ^ 

-  (3«„_,  +  ««-4  -  «»-3)«'.  T  (^^-.««-3  —  «»-ia«-4  -  3)z«'«-2 

Hence,  since  w, ,«/,,...  w^  are  positive  integers,  the  value  of 
;;/„  will  be  greatest  when  «/, ,  it', ,  w^ ,  7t\ ,  w, ,  .  .  .  are  greatest 
and  Ti'j,  zfc'j,  a;, ,  .  .  .  are  least  ;  and  the  value  of  »/,..,  will  be 
greatest  when  w„ ,  ze'„_, ,  w„_3,  w„_^,  .  .  .  are  greatest,  and  7i;'„_, , 
:i'„_^,  w„_<,,  .  .  .  are  least.  In  other  words,  the  bending  monr  ents 
at  the  1st  and  («  —  i)th  intermediate  supports  have  their  maxi- 


f 

it  i: 


1^ 


lit.     I 
Iritl'    ' 


[  I 


1l 


1- 


i 


^>  :jJ 


I 


III 


i  .!  .,i  if 


478 


THEORY  OF  STRUCTURES. 


mum  values  when  the  two  spans  adjacent  to  the  support  in 
question,  and  then  every  alternate  span,  are  loaded,  and  the  re- 
maining spans  unloaded. 

w, ,;«,,...  w„_j  may  now  be  easily  determined. 

Thus,  by  eq.  (i), 

P 

4 

/'  (  4 

•        +  (a!„_,«„_,  —  <j;«_,a„_3  —  i)«',  —  •  •  •  j 

+  («'„_,  —  I  — 4a,_,a„_,  +  4'^«-.««-3  +  4)^a  +•••}■ 

But  «,_,  =  4«„_j  —  dn-y 

4-  (3««-.««-3  +  3)«',  +  •  •  •  I . 

and  is  greatest  when  w, ,  ty, ,  w, ,  w, , .  .  .  are  greatest  and  a^, , 
7^4 ,  w, ,  w, ,  .  .  .  are  least. 

Similarly,  by  eqs,  (i)  and  (2), 

•    /•  /• 

w,  =  -  -  (w.  +  w.)  +  -  .  4(w,  +  w^  4-  15W, 
4  4 

=  -  -i ;  j  K-4<*«-.  +  4)Wi  -  (3««-.^»-4  +  I2)w, 


H-(ii««-i««-4  +  44K+  •  • 


CONTINUOUS  GIRDERS. 


479 


and  is  greatest  when  w^,w^,w^,w^,w^,  .  .  .  are  greatest  and 
w, ,  ly^ ,  w, ,  7fg ,  .  .  .  are  least. 

Thus,  the  general  principle  may  be  enunciated,  that  "in  a 
horizontal  continuous  girder  of  n  equal  spans,  with  its  ends 
resting  upon  two  abutments,  the  bending  moment  at  an  inter- 
mediate support  is  greatest  when  the  two  spans  adjacent  to 
such  support,  and  the  alternate  spans  counting  in  both  direc- 
tions, carry  uniformly  distributed  loads,  the  remainder  of  the 
spans  being  unloaded." 

Cask  II.  The  principle  deduced  in  Case  I  also  holds  true 
when  the  loads  are  distributed  in  any  arbitrary  manner. 

Consider  the  effect  of  a  weight  xv  in  the  rth  span  concen- 
trated at  a  point  distant/  from  the  {r  —  i)th  support. 

By  the  Theorem  of  Three  Moments, 

4w,  -f  w,  =  o ; (i) 

m^-\-^ni^-\rtn^  —  o\ (2) 

*«a  +  4w«.  4- '««  =  o ; (3) 


;«,.,  4-  4W,..,  -|-  nir  — 
;«,.,  +  4W,.  -f  vi,^,  = 


2/ 

M/-^(/' —/)= —yi,  suppose;    (r  —  i) 


=  —  w  -t(/  —  p){2l  —p)=—B,  suppose  ;  (r) 


»«,, -f  4w,+,  +  w,+,  =  o  ;       .    ,    .     .    (r-j-  i) 


w„.3  +  4w,.,  +  w„.,  =  o ;      ..,.(«  —  2) 
««ii-i  +  4w,-,  =0 («  —  0 


rtr-? 


'I 


■ 


m 


480  THEORY  OF  STRUCTURES. 

By  equations  (i),  (2),  (3),  ...  (r  —  2), 


II  I  I  w^  , 


the  upper  or  lower  sii^n  being  taken  according  as  r  is  even  or 
odd. 

By  equations  {n  —  i),  («  —  2),  («  —  3).  •  •  •  l''  +  0» 

w„_.  =  —  i;«H-2  =  tV'''«  3  ==  -  5*fl''^«-4  =  •  •  • 

=    +    =    ±   ~ —    -f •      . 

^H-r-i  "'H-r  "■H-r-l 

The  coeflficients  a  are  given  by  the  same  law  as  for  the  co- 
efficients a  in  Case  I.     Thus, 


a. 


«r-. 


w/^_,     and     w^^,  =  — 


(^H-r 


a 


nir . 


n-r+t 


Substituting  those  values  of  m^.^  and   m^+i  in  the  (r  —  i)th 
and  rth  equations, 


and 


where 


ri. 


w,_,^4  —  ^--'j  +  in,  =  —  A  =  m^.J}  +  m, 
-1  +  '"rU  -  T^^-J  =  —  B  =  nir.,  +  /«.<:, 


'«-f+2 


^  =  4  — and     tr  =  4  — . 

Hence,  solving  the  last  two  equations, 

Ac  -  B         ,                   Bb  —  A 
vtr-i^^ 1 and     w-= J . 

'^  be  —  I  '^  be  —  1 


The  ratios  -^'  and 


hence  b  and  ^  are  each  <  4  and  >  3 


are  each  less  than  unity,  and 


or 


I'L 


or 


I' I, 


or  the  co- 


COiV  TIN  UO  US   G/h'DKRS. 


481 


It  may  now  easily  be  shown  that  Ac  —  />  and  lib  —  A  arc 
each  positive.     Hence,  ///,._,  and  w,.  are  both  of  tiie  same  sii,ni. 

The  bendin^Lj  moment  ;//,,  at  any  intermediate  suijport  on 
the  left  of  /-  —  1  is  given  by 


III  -^  -j -III,.,  if  q  and  r  are  the  one  even  and  the  other  odd, 


or 


///,  — 


itr-l 


;;/,..,  if  q  and  r  are  both  even  or  both  odd. 


Thus  the  bending  moment  at  the  ^th  support  is  increased 
in  the  former  case  and  diminished  in  the  latter. 
If  (/is  on  the  right  of  r, 


III, 


or 


««- 


_|_  ^m±Lj,i^  if  q  and  r  are  both  even  or  both  odd, 


a 


tt-r+i 


n, 


III,—  — 


n-q+t 


ih,-r 


'--in^  if  q  and  r  are  the  one  even  and  the  other  odd, 


+2 


and  the  bending  moment  on  the  ^th  support  is  increased  in  the 
foiniLM-  case  and  diminished  in  the  latter. 

Thus,  the  general  principle  may  be  enunciated,  that,  "in  a 
hori/oiital  continuous  girder  of  n  equal  spans,  with  its  ends 
rcstiii<^  u[)on  two  abutments,  the  bending  moment  at  an  inter- 
nuuhate  support  is  greatest  when  the  two  spans  adjacent  to 
such  support,  and  the  alternate  spans  counting  in  both  direc- 
tions, are  loaded,  the  remainder  of  the  spans  being  unloaded." 

Case  III.  The  same  general  principle  still  holds  true  when 
the  two  end  spans  are  of  different  lengths. 

E.g.,  let  the  length  of  the  first  span  be  kl,  k  being  a 
numerical  coefficient,  and  let  2(1  -\- IS)  =  x, 

Eq.  (i)  now  becomes 


m^x  -(-  w,  =  o. 


: '■::*!# 


^%\ 


]■■' 
4  f 


:; 


482  THEORY  OF  STRUCTURES. 

Proceeding  as  before, 


Ill 


-  +  '"'  - 


/// 
7;, 


-'  =  +  ..., 


the  coefficients  /;, ,  /^j,  /',,  .  .  .  being  given  by  the  same  law  as 
before,  viz., 


/',   rz:    I   ; 

1>,=X', 

l\  =  4K  - 

-/\  =  4x-i; 

i\  =  aK  - 

-K=--  1 5-f  -  4 ; 

•         •         • 

-/;,  ^  56,t--  15; 

The  two  sets  of  coefficients  {a)  and  {b)  are  identical  when 
A-  =  4;  and  when  x  >  4,  all  the  coefficients  b  except  the  first 
(^,  =  i)  arc  numericallj'  increased. 

Hence,  the  same  general  results  will  follow. 

A'./>. — The  equations  giving  »i^  are  simple  and  easily  ap- 
plicable in  practice.     They  may  be  written 


and 


;//. 


in, 


(la      I^  ^t'    .r  .  y  \     r  r 

.  =  ±  — —  —. — - — -  It  ^  IS  on  the  left  of  r, 
'       "^  a,-,    be-  I       ^  ' 

=  ±   "«-i~i7~Z —  "  ^  '^  °"  '"^  ""'ght  of  r. 


If  there  are  several  weights  on  the  rth  span, 

A  =  2  -J?(/'  -  /)    and    B  =  ^7v  ^(/  -  p){2l  -  p). 

Example. — The  viaduct  over  the  Osse  consists  of  two  end 
spans,  each  of  94  ft.,  and  five  intermediate  spans,  each  of  126  ft. 
The  platform  is  carried  by  two  main  girders  which  arc  con- 


MAXlMUAf  BENDING  MOMENTS. 


483 


2;//,(/&+l)+'«,  =  -^V'.ii  +  i); 


4 


m,  -f  4;//,  4-  ;;/.  =  -  -(i|  +  i^) ; 


w,  +  4w.  +  m. 


/' 


di  +  i); 


*«6  +  4"^  +  "^  = 


4 


•      f«, +  2W,(/t+    l) 


4 


Hut  ^'  =  r^  =  f ,  very  nearly. 


^'    23 

7w,  +  2;«,  =  -  -  .  g- 


(0 


timious  from  end  to  end.  The  total  dead  load  upon  the  {girders 
may  l)e  taken  at  one  ton  (of  2000  lbs.)  per  lineal  foot. 

Denote  the  supports,  taken  in  order,  by  the  letters  a,  b,  c,  d, 
<'./•.'.''>  /''  •^"'^'  ''-"•^  't  '^^'  recpiired  to  find  the  maximum  bendinfj 
nioiueut  at  d  when  the  bridge  is  subjected  to  an  additional 
proc  '  load  of  l^  tons  [K-r  lineal  foot. 

The  spans  ab,  of,  dc,  /V  of  each  girder  carry  i^  tons  per 
lineal  f()ot. 

Tile  spans  be,  ef,  i^li  of  each  girder  carry  \  ton  per  lineal  foot. 

Denoting  the  bending  moments  at  a,  h,  c,  d,  c,  f,g,  h,  re- 
spectively, by  vi^ ,  ///, ,  .  .  .  w„,  the  intermediate  spans  by  /,  the 
end  spans  by  kl,  and  remembering  that  /«,  =  o  =  ?«, ,  we  have 


in. 


+  4"'3  +  '«4  =  -  7 


/'   7 


4    2 


(2) 


." ::  i 


484 


TIfKOKV   OF  STh'UCTUKKS. 

^'  5 
w,  +  4m,  4-  w,  =  -  -  .  J ; 


m^  +  4"^  +  w.  =  - 
w, -f  4/;/,  I- ;//,  =  - 

2w.  -1-  7;;/, 
From  cqs.  (i),  (2),  (3), 

97///,  -\-  26m ^  = 
From  cqs.  (4),  (5),  (6). 


4    2' 

^_:  7 

4'2' 

/•    '3 
4  ■  4  ■ 


(3) 

•    (4,1 


^  347 
4-   8  • 


26/;/,  -|-  gym,  = 


r  279 

4"   4  ' 
Hence,  m^,  the  niaxiimiin  required, 

/•        19151 


•4  ■  8  X  8733 


=  —  605.5  ft.-tons. 


16.  General  Theorem  of  Three  Moments. — The  most 
general  form  of  theorem  of  three  moments  may  be  deduced  as 
follows  : 

O, Oj  0, 


T^ 


I 


Fig.  334. 


Let  0,  X,  V,  the  (/•  —  i)th,  rth,  and  (r -f-  i)th  supports  of  a 
continuous  girder  of  several  spans,  be  depressed  the  vertical 
distances  U,  {=  0,0),  r/,  {=0,X),  and  ^/,  (=  0,V),  respectively, 
below  the  proper  level  0,0^0,  of  the  girder. 


ai-.iXi-.NAi.   riii'.OKi'.M  oi-  riiNEE  momi:.\'is.         485 

,1^  ,  </,, ,  r/,  aro  necessarily  very  small  (iiianlilics, 

III  (^CS  n\'  w  the  (UHleetion  curve,  and  let  the  lati^jeiit  at 

A'  nil  el   llu    verticals  llnoii^li    O  and    ('  in  /'.and  /'.and  the 

l,iii[,'enls  at  (^  and   V  \\\   7",  ami   '/',. 

Let  ^,  be  tlie  chaii^je  of  curvature  from  O  t(»  A'(=  0'I\E), 


li  11 


M  U 


"    /'to.r(^A7;K). 

1  et  ,  /,  ,  //,  be  the  t'Jfcclivc  moment  areas  for  the  spans  OX, 
\\\  respectively. 

Let  .\\  be  the  distance  hneasureil  horizontally;  of  the  centre 
(if  i;iavity  of  //,  from  O. 

\x\  .1,  be  the  distance  (measured  liorizontally)of  the  centre 
()( !.;ravity  of  //,  from  /''. 

Lrt  Oji  =j.,  OJ'^y,, 

Wy  Art.  2, 


/.  =  x,H,  = 


■i  —  -*i   I 


El 


'''. -j»  =  't'A  = 


Hut 


♦Hi  ^*i  »f(i    '  '''         »r  »r+l  ' 


7.   -  <    _  'K  -  )\  J.     ,       ^1  _  ^{»     ,      ^'a 


■' -s;^? 

1' 

■ 

-    F 

(    : 

n 

1;  ■  1 

and 


Again,  by  Art.  13,  Cor.  2, 

A,x,  =  y^,^",  +  \M,.j;  +  JiT/,// 


/Jr.  -^H,  being  the  areas  of  the  bcnding-moment  curves  for 
the  spans  OX,  XV,  respectively,  on  the  assumption  that  they 
are  independent  girders,  or  cut  at  0,  X,  and    V,  and  Sr,  ^r+i 


■,ij 


;h 


486 


THEORY  OF  STRUCTURES. 


being  the  horizontal  distances  of  the  centres  of  gravity  of  these 
areas  from  O  and  V.     Hence, 


7 


7 


+1 


/. 


+ 


S-+I 


iVi?/^. — If  0,X,ox  V  is  aiJ^i^^  ^,^3.  then  d^,  d^,  ord,'\s 
negative. 

Cor. — The  forms  of  the  Theorem  of  Three  Moments  given  in 
Cases  A  and  B,  Art.  13,  may  be  immediately  deduced  from  the 
last  equation. 

Case  A. 
d,  =0  =  d,=d,; 


^"-3     8     ^'■' 


/. 


^r  =  ~; 


...  _  6A,/  -  eA^^.-p  =  ~  \{wj;  +  «^.+,/'h-.). 


Case  B. 
d,=o  =  d.,  —  d,\ 


Ir+P 


17.  Advantages  and  Disadvantages  of  Continuous 
Girders. — The  advantages  claimed  for  continuous  girders  are 
facility  of  erection,  a  saving  in  the  flange  material,  and  the  re- 
moval of  a  portion  of  the  weight  from  the  centre  of  a  span  to- 


''t 


w 


is 


PKOPERTIES  OF  CONTINUOUS  GIRDERS. 


487 


wards  the  piers.  Circumstances,  however,  may  modify  these 
advantages,  and  even  render  tliem  completely  valueless.  The 
fliuifre  stresses  are  governed  by  the  position  of  the  points  of  in- 
tlcxion,  which,  under  a  moving  load,  will  fluctuate  through  a 
distance  dependent  upon  the  number  of  intermediate  sujjports 
aiul  upon  the  nature  of  the  loading.  In  bridges  in  which  the 
ratio  of  the  dead  load  to  the  live  load  is  small  the  fluctuation  is 
considerable,  so  that  for  a  sensible  length  of  the  main  girders, 
a  passing  train  will  subject  local  members  to  stresses  which  are 
alternately  positive  and  negative.  This  necessitates  a  local 
increase  of  material,  as  each  member  must  be  designed  to  bear 
a  much  higher  stress  than  if  it  were  strained  in  one  way  only. 

Again,  the  web  of  a  continuous  girder,  even  under  a  uni- 
formly distributed  dead  load,  is  theoretically  heavier  than  if 
each  span  were  independent,  and  its  weight  is  still  further  in- 
creased when  it  has  to  resist  the  complex  stresses  induced  by 


a  moving  load. 


Hence,  in  such  bridges  the  slight  saving,  if  there  be  any, 
camiot  be  said  to  counterbalance  the  extra  labor  of  calculation 
and  workmanship. 

In  girders  subjected  to  a  dead  load  only,  and  in  bridges  in 
which  the  ratio  of  the  dead  load  to  the  live  load  is  large,  the 
saving  becomes  more  marked,  and  increases  with  the  number 
of  intermediate  supports,  being  theoretically  a  maximum  when 
the  number  is  infinite.  This  maximum  economy  may  be  ap- 
proximated to  in  practice  by  making  the  end  spans  about  four- 
fifths  the  intermediate  spans. 

In  the  calculations  relating  to  the  Theorem  of  Three  Mo- 
ments, it  has  been  assumed  that  the  quantity  EI  is  constant, 
while  in  reality  E,  even  for  mild  steel,  may  vary  10  or  15  per 
cent  from  a  mean  value,  and  /  may  vary  still  more.  It  does 
not  appear,  however,  that  this  variation  has  any  appreciable 
effect  if  the  depth  of  the  girder  or  truss  Q\\'&wg<i^  gradually,  but 
the  effect  may  become  very  marked  with  a  rapid  change  of 
depth,  as,  e.g.,  in  the  case  of  swing-bridges  of  the  triangular 
type. 

The  graphical  method  of  treatment  may  still  be  employed 
by  substituting,  for   the  actual   curve   of   moments,  a  reduced 


\{ 


T 


i 


hi; 


Mi;' 


i 


488 


THEORY  OF  STRUCTURES. 


curve,  formed  by  changing  the  lengths  of  the  ordinates  in  the 
ratio  of  the  value  of  EI  a.t  a  datum  section  to  EI. 

It  is  often  found  economical  to  increase  the  depth  of  the 
girder  over  the  piers,  which  introduces  a  local  stiffness  and 
moves  the  points  of  inflexion  farther  from  the  supports.  A 
point  of  inflexion  may  be  made  to  travel  a  short  distance  by 
raising  or  depressing  one  of  the  supports. 

In  order  to  insure  the  full  advantage  of  continuity  the  ut- 
most care  and  skill  are  required  both  in  design  and  workman- 
sliip.  Allowance  has  to  be  made  for  the  excessive  expansion 
and  contraction  due  to  changes  of  temperature,  and  the  piers 
and  abutments  must  be  of  the  strongest  and  best  description 
so  that  there  shall  be  no  settlement.  Indceu,  the  difficulties 
and  uncertainties  to  be  dealt  with  in  the  construction  of  con- 
tinuous girders  are  of  such  a  serious  if  not  insurmountable 
character  that  American  engineers  have  almost  entirely  dis- 
carded their  use  except  for  draw-spans. 

Much,  in  fact,  is  mere  guesswork,  and  it  is  usual  in  prac- 
tice to  be  guided  by  experience,  which  confines  the  points  of 
inflexion  within  certain  safe  limits. 

Under  these  circumstances  it  may  prove  desirable  to  fix 
the  points  of  inflexion  absolutely,  and  the  advantages  of  doing 
so  are  (a)  that  the  calculation  of  the  web  stresses  becomes 
easy  and  definite,  instead  of  being  complicated  and  even  in- 
determinate ;  {b)  that  reversed  stresses  (for  which  pin-trusses 
are  less  adapted  than  riveted  trusses)  are  almost  entirely 
avoided ;  {c)  that  the  stresses  are  not  sensibly  affected  by- 
slight  inequalities  in  the  levels  of  the  supports;  {d)  that  the 
straining  due  to  a  change  of  temperature  takes  place  under 
more  favorable  conditions. 

The  fixing  may  be  thus  effected  : 

{a)  A  hinge  may  be  introduced  at  the  selected  point. 

The  benefit  of  doing  so  is  very  obvious  when  circumstances 
require  a  wide  centre  span  and  two  short  side  spans. 

{b)  If  the  web  is  open,  i.e.,  lattice-work,  the  point  of  inflex- 
ion in  the  upper  flange  may  be  fixed  by  cutting  the  flange  at 
the  selected  point  and  lowering  one  of  the  supports  so  as  to 
produce  a   slight    opening   between  the  severed   parts.     The 


ADVANTAGES   OF  CONTINUOUS  GIKDERS. 


489 


position  of  the  point  of  inflexion  in  the  lower  flange  is  then 
defined  by  tlie  condition  that  the  algebraic  sum  of  the  hori- 
zontal components  of  the  stresses  in  the  diagonals  intersected 
by  a  line  joining  the  two  points  of  inllexion  is  zero. 

It  must  be  remembered,  however,  that  \.\V\ii  Jixiiig  of  the 
points  of  inflexion,  or  the  cutting  of  the  chords,  destroys  the 
property  of  continuity,  and,  indeed,  is  the  essential  distinction 
between  a  continuous  girder  and  a  cantilever. 

I'our  methods  may  be  followed  in  the  erection  of  a  contin- 
uous girder,  viz. : 

1.  It  may  be  built  on  the  ground  and  lifted  into  place. 

2.  It  may  be  built  on  the  ground  and  rolled  endwise  over 
the  piers.  As  the  bridge  is  pushed  forward,  the  forward  end 
acts  as  a  cantilever  for  the  whole  length  of  a  span,  until  the 
next  pier  is  reached.  This  method  of  erection  is  common  in 
Fiance. 

3.  It  may  be  built  in  position  on  a  scaffold. 

4.  Each  span  may  be  erected  separately,  and  continuity  pro- 
duced  by  securely  jointing  consecutive  ends,  having  drawn  to- 
gether the  upper  flanges.  A  more  effective  distribution  of  the 
material  is  often  made  by  leaving  a  little  space  between  the 
flanges  and  forming  a  wedge-shaped  joint. 


w? 

f'n 

\ 

31 


i: 

i  '■: 
I 
*■  i " 


490 


THEORY  OF  sr/ii/CTUNES. 


EXAMPLES. 

1.  Two  angle-irons,  each  2  in.  x  2  in.  x  i  in.,  were  placed  upon  sup 
ports  12  ft.  9  in.  apart,  the  transverse  outside  distance  between  the  bars 
being  9i  in.,  and  were  prevented  from  turning  inwards  by  a  thin  phuo 
upon  the  upper  faces.  Tlie  bars  were  tested  under  uniforndy  distributed 
loads,  and  each  was  found  to  have  deflccied  2,\^  in.  wiien  tiie  load  over 
the  two  was  1008  lbs.     Find  A"  iiiid  the  position  of  the  neutral  axis. 

Ans.  /=  ,V/4  ;    E=  17,226,139   lbs.;    neutral  axis  J^  in.  from 
upper  face. 

2.  Botii  bars  in  the  preceding  question  failed  together  when  the 
total  load  consisted  of  loi  cwts.  (cwt.  =  112  lbs.)  uniformly  distributed, 
and  3  cwts.  at  the  centre.     Find  the  maximum  stress  in  the  metal. 

Ans.  Compressive  unit  stress  =  20,323  lbs.; 
Tensile  unit  stress  =  39,577  lbs. 

3.  Show  that  the  moments  of  resistance  of  an  elliptic  section  and  of 
the  strongest  rectangular  section  that  can  be  cut  out  of  the  same  are  in 
the  ratio  of  99  V3  to  112,  and  that  the  areas  of  the  sections  are  in  the 
ratio  of  33  to  14  \'2. 

4.  Show  that  the  moments  of  resistance  of  an  isosceles  triangular 
section  and  of  the  strongest  rectangular  section  that  can  be  cut  dut  uf 
the  same  are  in  the  ratio  of  27  to  16,  and  that  the  areas  of  tiie  twu 
sections  are  in  the  ratio  of  9  to  4. 

5.  An  angle-iron,  3  in.  x  3  in.  x  ^^^  in.,  was  placed  upon  supports 
12  ft.  9  in.  apart,  and  deflected  ij  in.  under  a  load  of  8  cwts.  uniformly 
distributed  and  2  cwts.  at  the  centre,  Find  E  and  the  position  of  the 
neutral  axis. 

Ans.  E  —  16,079,611  lbs.;  neural  axis  \\%  in.  from  upper  face, 

6.  The  effective  length  and  cenuai  depth  of  a  cast-iron  girder restin;,' 
upon  two  supports  were  respectively  1 1  ft.  7  in.  and  10  in. ;  the  bottom 
flange  was  10  in.  wide  and  \\  in.  thick;  the  top  flange  was  2^  in.  wide 
and  J  in.  thick;  the  thickness  of  the  web  was  \  in.  The  girder  was 
tested  by  being  loaded  at  points  3!  ft.  from  each  end,  and  failed  when 
the  load  at  each  point  was  ^^\  tons.  What  were  the  total  central 
flange  stresses  at  the  moment  of  rupture .' 

What  was  the  central  deflection  when  the  load  at  each  point  was  7i 
tons?     {E  =  18,000,000  lbs.,  and  the  weight  of  the  girder  —  3368  lbs.) 

Alts.    164,747.4  lbs. ;  .36S  in. 


mmmm^ 


EXAMPLES. 


491 


7.  A  tubular  ^jirrler  rests  upon  supports  36  ft.  apart.  At  6  ft.  from 
otif  011(1  the  tlaiijjfs  arr  cadi  27  in.  wick-  arid  2%  in.  thick,  the  net  area  (jf 
the  tension  t1an>re  beinj^  60  in.,  whih  the  web  consists  of  tv/o  ^f'\n. 
plaus,  j6  in.  clet'i)  and  18  in.  apart.  V  glectin^'  the  elfect  of  the  anj^le- 
iroris  uniting  tin:  web  plates  to  the  llanges,  determine  tlie  tiuiinent  of 
resistance. 

The  (,'irder  has  to  carry  a  uniformly  distributed  dead  loail  of  56  tons, 
a  uniformly  distributed  live  load  <'f  54  tons,  and  a  local  load  at  the 
j^'ivin  section  of  100  t(jns.  What  are  the  corresponding  (lange  stresses 
per  s(|uare  inch  } 

How  many  J-in.  rivets  are  required  at  the  given  .section  to  unite  the 
angle-irons  to  the  llanges? 

Ans.  238.13  X  coeff.  of  strength  ;  3.3186  tons  ;  3.896  tons. 

8.  A  yellow-pine  beam,  14  in.  wide  and  15  in.  deep,  was  placed  upon 
supports  10  ft.  9  in.  apart,  and  deflected  j  in.  under  a  load  of  20  tons  at 
the  centre.     Find  K,  neglecting  the  weight  of  the  beam. 

.Ills.  11  =  1,272,112  lbs. 

9.  What  were  the  intensities  of  the  normal  and  tangential  stresses  at 
:  ft  from  a  sujjport  and  2.i  in.  from  neutral  jjlane,  upon  a  plane  inclined 
at  30"  to  the  axis  of  the  beam  in  the  preceding  question.-' 

A/is.   132.83  and  218.91  lbs. 

10.  A  beam  is  supported  at  the  ends  and  bends  under  its  own  weight. 
Shoa-  that  the  upward  force  at  the  centre  which  will  exactly  neutralize 
llic  bending  action  is  equal  to  ^  or  J  of  the  weight  of  the  beam  (u'), 
according  as  the  ends  avi^/rtY  ox  fixed. 

Find  the  neutralizing  forces  at  the  quarter  spans. 

Ans.  Ends  free   ^'^y,io  at  eaeli  or  ,'."f,7i/ at  one  of  the  points  of 
divisif)n. 
Ends  fixed  -^^70   at   eue/i  or   ^70  at  one  of    the  points  of 
division. 

11.  A  beam  8  in.  wide  and  weighing  50  lbs.  per  cubic  foot  rests  upon 
supports  30  ft.  apart.  Find  its  depth  so  that  it  may  deflect  |  in.  unrier 
its  own  weight.     (E—  i  ,200,000  lbs.)  Ans.  9.185  in. 

12.  A  rectangular  girder  of  given  length  (/)  and  breadth  (/>)  rests 
upon  two  supports  and  carries  a  weight  /'  at  the  centre.  Find  its  depth 
so  that  the  elongation  of  the  lowest  fibres  may  be  ,5'^nj  "^  ^'"^  original 
length. 

/2100/'/ 


Ans.  \/ . 


oh 


13.  A  yellow-pine  beam,  14  in.  wide,  15  in.  deep,  and  weighing  32  lbs. 
per  cubic  foot,  was  placed  upon  supports  10  ft.  6  in.  apart.  Under 
uniformly  distributed  loads  of  59,734  lbs.  and  of  127,606  lbs.  the  central 


V-.}- 


492 


THEORY  or  STKUCTURF.S. 


I    f 


dencctions  were  respectively  .uS  in.  .iiul  .29  in.     FMnrl  the  mean  value 

of  /•;. 

Also  (letcrinine  the  additional  weii^lit  at  the  cx-ntre  which  will  increase 
the  lirst  dcHection  by  ,'0  of  an  inch.        Atis.   2,552,980  lbs.;  24,121  lbs. 

14.  In  tile  preceding  question  lind  for  the  load  of  59.734  lbs.  ilie 
inaximuni  intensities  of  thrust,  tension,  and  shear  at  a  point  half  wav 
between  the  neutral  axis  and  the  outside  skin  in  a  transverse  set  tioii  ,ii 
one  of  the  points  of  trisection  of  the  beam.  Also  find  the  inclinaliniis 
of  the  planes  of  principal  stress  at  the  point. 

Alts.   1O09.255,  165.562,  1 19.364  lbs. ;  0  =:  3"  48f. 

15.  A  pitch  pine  beam,  14  in.  wide,  15  in.  deep,  and  weighing  45  lbs. 
per  cul)ic  foot,  is  jjlaced  upon  supports  10  ft.  9  in.  apart,  and  can  ies  ,1 
load  of  20  tons  at  the  centre.  Find  the  deflection  and  curvature,  /; 
being  1,270,000  lbs.     What  slitTness  does  this  give  } 

Wliat  amount  of  uniformly  distributed  load  will  produce  the  same 
dellection  ?  Ans,  ^\^\  32  tons. 

16.  In  the  preceding  question  find  the  maximum  intensities  of  thrust. 
tension,  and  shear  at  points  (a)  half-way  between  the  neutral  a.\is  and 
the  outside  skin,  (/')  at  one  third  of  the  depth  of  the  beam,  in  a  lians- 
vcrse  section  at  one  of  the  quarter  spans.  Also  lind  the  inclinations  of 
the  planes  of  principal  stress  at  tiiese  points. 

Ans.  — {a')  951. S53,  292.969,  329.442  lbs.;  0=9°  34*'. 
{!>)  65S.774,  171.108,  243. S33  lbs.;  0^  r5    50'!'. 

17.  A  piece  of  greenheart,  142  in.  between  supports,  9  in.  dcei).  ,mil 
5  in.  wide,  was  tested  by  being  loaded  at  two  points,  distant  23  in.  from 
the  centre,  with  equal  weights.  Under  weights  at  each  point  of  44^10 
lbs.,  11,200  lbs.,  and  17,920  lbs.  the  central  deflections  were  .  13  in..  .37 
in.,  .67  in.,  respectively.  Find  the  mean  coelFicient  of  elasticity.  The 
beam  broke  under  a  load  of  32,368  lbs.  at  each  point.  Find  the 
coefHcient  of  bending  strength. 

18.  A  sample  cast-iron  girder  for  the  Waterloo  Corn  Warehouses. 
Liverpool,  20  ft.  J\  in.  in  length  and  21  in.  in  depth  (total)  at  the  centre, 
was  placed  upon  supports  iS  ft.  \\  in.  apart,  and  tested  under  a 
uniformly  distributed  load.  The  to|)  flange  was  5  in.  x  i^  m.,  the 
bottom  tlange  was  18  in.  x  2  in.,  and  the  web  was  \\  in.  thick.  The 
girder  deflected  .15  in.,  .2  in.,  .25  in.,  and  .28  in.  under  loads  (including' 
weight  of  girder)  of  63,76311)3.,  88,571  lbs.,  107,468  lbs.,  and  119,746  lbs., 
respectively,  and  broke  during  a  sharp  frost  untler  a  load  of  390,282  li)s. 
Find  the  mean  coelTicient  of  elasticity  and  the  central  flange  stresses  at 
the  moment  of  rupture. 

Am.  7  =  3309.122;  £■  =  17,427,327  lbs.;  20,121  lbs.,  47,168  lbs. 

19.  A  steel  rectangidar  girder,  2  in.  wide,  4  in.  deep,  is  placed  upon 


EXAA/J'LEU, 


493 


supports  20  ft.  apart.     If  K  is  35,ckx),ooo  Ihs.,  fiiul  the  woij4ht  wliicli,  il 
placed  at  tlic  centre,  will  rau,se  the  beam  to  dellect  i  in. 

j-lns.    I  stX^sV  't^s. 

20.  A  timber  joist  weif^iiinfj  48  lbs  per  cubic  foot,  2  in.  wide  x  12  in, 
deep  X  14  ft.  long,  deflected  .825  in.  iirder  a  load  of  887  ll)s.  at  tiic 
cciiire.     Find  A'.  Ans.  3(^7,880  lbs. 

21.  A  beam  of  span  /  is  uniformly  Jdaded.  Cnmpare  its  strenglii  and 
sliflness  (rt)  when  merely  resting  upon  supports  at  the  ends;  (/')  vvlu'ii 
fixed  at  one  end  and  resting  upon  a  support  at  the  otlur;  (i)  when  tix(;(l 
at  both  ends.  In  case  ic)  two  liingi.'s  an;  introfluced  at  points  distatit  y 
(roin  the  centre  ;  show  that  the  s/rcfii^/Zi  oi  the  beam   is  economized  to 

tlie  best  ellect  when/  —  -       .  and  Uiat  the  stiJTness  is  a  maximum  when 

252 

y  =  -  very  nearly. 
4 

Ans.  Cases  {a)  and  (h).     iitx  :  nix  ::  i  :  i  ;     D\  .  Di  ::  i  :  .416. 
Cases  (a)  and  (< ).     ni\  :  Wa  ::  3  :  2  ;     l)\  :  Ih  '•'•  10  :  3. 
Case    ((•).  Max,  economy,  Wi  :  im  ::  2  :  i  : 

1),  .  /A  ::  5  :  2  y'2. 
Max,   stiffness,  Wi  :  ///;,  ::  4  :  3  ; 

/^i  :  Dt,  ::  15:4  (approx,). 

22.  A  beam  AB  o[  span  /,  carrying  a  uniformly  distributed  load  of 

intensity  w,  rests  upon  a  support  at  //  and   is  imperfectly  fixed  at  „■/,  so 

I    7.'/' 
tliat  the  neutral  axis  at  ,/   has  a  slojjc  of 


tlian  .  /  f)y  an  amount 


I    ivl' 


48  A7  ■ 
Find  the  reactions, 


The  end  //  is  lower 


How  much  must  li 
Fmri 


Ans    ■     -joi,  — <f/ ; 
3^        32 


32  I'J  ■ 
he  luwered  so  that  the  whole  of  the  weight  may  be  l)orne  at  // 
the  work  done  in  bending  the  beam. 

7   «'/♦ 
48  Kl  • 

23.  .A  round  wrought-iron  bar  /  ft,  long  and  (/in.  m  diameter  can 
just  carry  its  own  weight.  Find  /  in  terms  of  d,  (n)  the  allowable  dc- 
flecti'in  being  i  in,  per  100  ft,  of  span,  A"  being  30,000,000  lbs,;  {/n  the 
allowable  stress  being  8960  lbs.  per  square  inch  ;  (0  the  stilTness  given  bv 
(iM  and  the  strength  given  by  [h)  being  of  equal  im|)ortance, 

Ans.     [a)  I  ^  yzipiF'  ;    (h)  I  -  ^224^  ;    <.■:)  /=  Wd. 
i\.  A  square  steel  bar  i  ft.  long  and  having  a  side  of  length  //  in.  can 
just  carry  its  own  weight;  its  stiffness  is  i^',,^  and  the  maximum  allow- 
able working  stress  is  7  ions  per  square  inch.     Find  /  in  terms  of  (/,  E 


being  13,000  tons. 


fins. 


/fin  ft.) 
d {\\\  in.) 


13 
7' 


%l 


p 


!,:■    1 


^ 


494 


THEORY  OF   STRUCTURES. 


25.  A  uniformly  loaded  beam  with  both  ends  absolutely  fixed  is 
hinged  at  the  quarter-spans.  Show  that  the  slope  is  suddenly  doubled 
on  passing  a  hinge. 

26.  A  horizontal  beam  with  both  ends  absolutely  fixed  is  loaded  u  iiii 
a  weight  W  at  a  point  dividing  the  span  into  two  segments  a  and  /;, 

Show  tliat   the   deflection  at  the  point  is 


3AY  \a  ^-  h 


■work  done  in  bending  the  beam. 


Ans. 


(iEI 


and  find  the 

~ii  +  b 

27.  Determine  the  isosceles  section  of  maximum  strength  which  can 
be  cut  out  of  a  circular  section  of  given  diameter,  and  compare  the 
strengths  of  the  two  sections. 

28.  A  3-in.  X  3-in.  X  ^-in.  angle-iron,  with  both  ends  fixed  and  a  clear 
span  of  20  ft.,  carries  a  uniformly  distributed  load  of  500  lbs.  which  causes 
It  to  deflect  2  in.  Find  E.  What  single  load  at  the  centre  will  produce 
the  same  deflection  }     Find  the  work  done  due  to  bending  in  each  case. 

Ans.  E  =  20,775,415  lbs.;  250  lbs. 

29.  A  steel  plate  beam  of  uniform  section  and  30  ft.  span  has  belli 
ends  fixed  and  is  freely  hinjied  at  the  points  of  triscction.  Doteniiiiie 
the  neutral  axis(«)  for  a  uniformly  distributed  load  of  60,000  lbs.;  (/')  for 
a  single  load  of  10,000  lbs.  concentrated,  ^a-j/,  7^  ft.  and,  second,  15  ft. 
from  a  support. 


Ans.  (ci)  bide   span,/  =  — 

EI 


100 


S-v  + 


1750000  ,  25.1- 
centre  span,/  =  -^^,-,-j-  -f-  -V> 
3A/         3A/ 


loo  —  2o.r''  +  .r'j. 
(b)  First.   Loaded  span  between  support  and  weight. 


)' 


;oo  /      „       \ox\ 


Loaded  span  between  weight  and  hinge, 

y  = -—f  28i25o.r  -  703125  j. 

Unloaded   side   span    horizontal ;    centre  span 
straight  between  liinges. 


5000000 


5ooo.r/  x2 

Second.  Side  span,  v  =  -  >rr    5-»'  —  -7- 

hl    \  6 

2500.V/  .f2\ 

centre  b^an,/  =  ->y-(25 ]  + 


\L^ 


jntre  span 


EXAMPLES. 


495 


30.  A  uniformly  loaded  beam,  with  both  ends  absolutely  fixed,  is 
hinged  at  a  point  dividing  the  span  into  segments  a  and  b.  Draw  curves 
of  siiearing  force  and  bending  moment,  and  compare  the  strength  and 
stirtness  of  the  beam  when  the  hinge  is  {a)  at  the  middle  point  ;  {b)  at 
a  point  of  trisection  ;  yc)  at  a  quarter-span.  Also,  determine  the  slope  of 
the  segments  of  these  points. 

IV  ^a"  +  8rt^'  +  ib^ 


Ans.    A'l  = 


<«•  +  b' 


K: 


loa  a*  +  4<?<*°  +  Zb* 
8  a*  +  b^ 


M'  -.A/"  :  J/'"::  14:  14:  11;   D' 
I  fx 


_  w  3rt*  +  ^a^b  +  5^« 
~  8  rt»  +  b^  ' 

Af  _  ^''  3«*  +  Aa^b  +  b* . 
8  rt'  +  /)' 

D'  :/>"'::  6.25  :  3.29  : 


2.66. 


1/2 


Slopes  in  {a)  =  —  —  -  ;  in  (^)  =  —  -  —    for  segment  a, 
E  c  b  h  ct>\ 

/  /  23 
and  =  —  —  ---  for  segment  b ; 
t  f  162 

in  (c)  =.  —  —  -~  for  segment  a,  and 
^ '^      E  c  176  ^ 

/   /  92 
»  i:  —  ^  —  for  segment  b. 

E  c  2,g\  ^ 

31.  A  horizontal  beam  rests  upon  two  supports  and  is  loaded  with  a 
weight  fF at  a  point  dividing  the  span  into  segments  a  and  b.  Find 
the  deflection  at  this  point  and  the  worli  done  in  bending  the  beam. 


Ans. 


IV       a-'b* 


IV    a'b' 


3  E/[a  +  b) '  bEI  a  +  b 


—  —  X  deflection 


32.  A  wrought-iron  beam  of  rectangular  settion  and  20  ft.  span  is 
16  in.  deep,  4  in.  wide,  and  is  loaded  with  o  proof  load  at  the  centre.  If 
the  proof  strength  is  7  tons  per  square  inch,  find  the  proof  deflection  and 
the  resilience,  E  being  12,000  tons.  Arts.  .029  ft.  ;  650  ft. -lbs. 

33.  Design  a  wooden  cantilever  12  ft.  long,  of  circular  section  and 
uniiorm  strengtli,  to  carry  a  uniformly  distributed  load  of  2  tons,  the 
(oetficitMit  of  working  strength  being  i  ton  per  square  inch.  Alst),  find 
the  deflection  of  the  free  end. 

Ans.  Taking  fixed  end  as  origin  and  3  being  radius  in  inches  at 

distance  x  ft.  from  origin,  then  i  ic''  =  141 12  —  .r)'. 

_   „      .  ,       697.6 

Deflection  at  end  = m. 

E 

34.  A  girder  fixed  at  both  ends  carries  (2«  4-  i)  weights  ff  concen- 
trated at  points  dividing  the  length  of  the  girder  into  2/1  +  2  equal 
divisions.     Find  the  total  central  deflection.  ,       «  +  i  IV/* 


Ans. 


192     EI 


496 


THEORY  OF  STRUCTURES. 


!.i    '.) 


i:  1 


35.  A  girder  30  m.  long  has  both  ends  rixed  and  carries  a  uniformly 
distributed  load  of  5800  k.  per  lineal  metre.     Find  the  deflection  and 


the  work  of  flexure. 


^„,.  &21^ooo  ^_^^ 


36.  A  steel  beam  of  circular  section  is  to  cross  a  span  of  15  ft.  and 
to  carry  a  load  of  10  tons  at  5  ft.  from  one  end.  Find  its  diameter,  the 
stiffness  being  such  that  the  ratio  of  viaximutn  deflection  to  span  is 
.00125.     £=  1 3,000  tons.  Ans.   10.3  ui. 

37.  Determine  the  dimensions  of  a  beam  of  rectangular  section 
which  might  be  substituted  for  tiie  round  beam  in  the  preceding  ques- 
tion, the  stiffness  remaining  the  same  and  the  coefficient  of  working 
strength  being  7i  tons  per  square  inch.  Ans.  lui^  —  320. 

38.  The  flange  of  a  girder  consists  of  a  pair  of  angle-irons  and  of  a 
plate  which  extends  over  the  middle  portion  of  the  girder  for  a  cci  tain 
required  distance.  Show  that  the  greatest  economy  of  materia!  is 
secured  when  the  length  of  the  plate  is  t  of  the  span  and  the  sectional 
areas  of  the  plate  and  angle-irons  are  as  4  to  5  (the  girder  being 
uniformly  loaded). 

39.  The  flange  of  a  uniformly  loaded  girder  is  to  consist  of  two 
plates,  each  of  which  extends  over  the  middle  portion  of  the  girder  fur 
a  certain  required  distance,  and  of  a  pair  of  angle-irons.  Show  tiiat 
the  greatest  economy  of  material  is  realized  when  the  lengths  of  tlie 
plates  and  angle-irons  are  in  the  ratio  of  12:18:  23,  and  when  the  areas 
of  the  plates  are  in  the  ratio  of  4  :  5. 

What  should  be  the  relative  lengths  of  the  plates  if  they  are  of  equal 
sectional  area?  Ans.    i  •.\'lUS2-\-  \). 

40.  An  elastic  beam  rests  upon  supports  at  its  ends,  and  a  weight 
placed  at  a  point  A  produces  a  certain  deflection  (d)  at  a  point  B. 
Show  that  if  the  weight  is  transferred  to  B  the  same  deflecti(jii  (V/i  is 
produced  at  A. 

41.  A  uniform  beam  is  supported  by  four  equidistant  props,  of  which 
two  are  terminal.  Show  that  the  two  points  of  inflexion  in  the  middle 
segment  are  in  the  same  horizontal  plane  as  the  props. 

42.  Find  the  slope  and  deflection  at  the  free  end  of  the  following 
cantilevers  when  bending  under  their  own  weight,  /  being  the  length, 
2b  the  depth  at  the  fi.xed  end,  10  the  specific  weight,  and  £■  the  coeMicient 
of  elasticity : 

{a)  Of  constant  thickness  /  and  with  profile  in  the  form  of  a  trapezoid 
with  the  non-parallel  sides  equal  and  of  depth  2a  at  the  free  end. 

{b)  Of  circular  section  and  with  profile  in  the  form  of  an  isosceles 
triangle. 


nifornily 
lion   and 

—  km. 

15  11.  and 
leter,  the 
3  span  is 
10.3  in. 

xr  section 
[ins;  ques- 
f  working 

"  =  3-0- 

5  and  of  a 
r  a  certain 
material  is 
e  sectional 
irder  beini; 

isist  of  !\\o 

J  girder  t'C 

Show  tiiat 

;ths  of  liie 

en  the  areas 

ire  of  equal 

|nd  a  weight 

a   point  />'. 

Iction  ((/>  i^ 


js.  of  wliicli 
the  middle 


lie  following 

the  Icniitli, 

|e  coefficient 

a  trapezoid 
end. 


Lin   i#oscc 


les 


EXAMPLES. 


497 


((•;  Of  constant  thickness  and  with  profile  in  the  form  of  a  parabola 
symmetrical  with  respect  to  the  ;ixis  and  having  its  vertex  at  the  fiee 
end. 


z^'P  3?c'/        \ 

AflS.—  (0-    rr-r;    -.-.r       ""si 


lOg  ,  f  


I       -UJi,' 


'^^rr,. 


'£' 


('-■) 


b 


6(5' 


43.  Deduce  the  slope  and  deflection  at  the  free  end  — 

((/)  When  the  depth  2ii  in  {a)  of  the  preceding  question  is  «//,  i.e., 
when  the  profile  is  an  isosceles  triangle. 

(ij  Due  to  a  unifcjrmly  distributed  load  of  intensity  p  over  the 
cantilever  ((?).  Hence,  also,  deduce  the  slope  and  defiection  when  the 
depth  2ii  is  /!/'/. 

{/  \  Due  to  a  wcitiht  H'at  the  free  end  of  {a). 

(,!,'i  Due  to  a  uniformly  distributed  load   of   intensity  p  upon   the 

cantilever  0). 

U'P       UT 
AHs.—((i) 


(^) 


2/3''     .\Eb'' 

3  pi'        \  ,       />        (ib  -  a)(b  -  a) 

4  EtKb  —  ay'  (      '    a  2b'' 


pP        \ 


4  £/(b  ■ 

3  /{l 

4  A7^'  ■ 


_;-.,3.'l<.g-^~  + 


a       (2^+  lab  -  a'')(b  -  a)  ) 


(g) 


p     P 


Eb't ' 


3  _frp 

2  E/ib- 

10  aW 


(1      ^ 

)  log- 
— r.  "\  a 


2b'' 


jb  —  a 
2b' 


\ 


90" 


44..  A  cantilever  of  length  /,  specific  weight  w,  and  square  in  section, 
aside  of  tiie  section  being  2b  at  the  fixed  and  2a  at  the  free  end,  bends 
under  its  own  weitiht.  Find  the  slope  and  deflection  of  the  neutral  axis 
at  the  free  end.  Hence,  also,  deduce  corresponding  results  when  the 
cantilever  is  a  regular  pyramid. 

(/;  +  i^yi>P_  _    {b  +  2a\n<n       wP       wl' 
AEP       '  8Eb'        '  ^Eb" '  JEb''' 

45.  If  the  section  of  the  cantilever  in  the  preceding  question,  instead 
of  being  square,  is  a  regular  figure  with  a)ij>  number  of  equal  sides,  show 
that  the  neutral  axis  is  a  parabola  with  its  vertex  at  the  point  of  fixture. 

4^^'.  The  section  of  a  cantilever  of  length  /  is  an  ellipse,  the  major  axis 
(verihii/)  being  twice  the  minor  axis.     Find  the  deflectfon  at  the  end 


a 


m 


iti 


\ 


498 


r/f/':oA'y  or  sjw^uctu/^ks. 


ntiflor  II   hiiij^Ii"  wciglit    //',/  Ijciii^;  llu"  cociricicnt   nf   wotkiiiii;  stic  iii;ili 
and  A' tin' liU-HuioiU  of  cliisticity.  / -M7      /V''   \\ 

■■'"'■  [7000 /-ir)- 

47.  A  cast-iioii  IxMin  i>f  an  iiivottoii  'rseclioii  rests  upon  sii|iiinits 
22  ft.  apart  ;  tlu-  wcl)  is  i  in.  iliicU  and  20  in.  doep;  tlu-  tlan^i-  is  1,:  m. 
tliirk  and  u  in.  widi-;  ilic  Ix'ain  laiiii-s  a  unifoindy  (iislrihuti'd  Icjad  di 
99,ixx)  lbs.      I'liid  tlu'  nia.xiiniim  dutk-clion,  A' i)i'inj;  ij.ijjo.ckxj  lbs. 

,  ht.s.  .Mj  J  in.  (  /  —  KjdS.o^i. 
4S.    I'ind  the  ni.ixiinnni  dcfk-ction  of  a  casi-iion  canliliAcr  2  1.1.  wkIc 
X  3  in.  deep  x   1:0  in.  long  under  its  own  wriglit.  A,"  being  17,920.000  llis. 

.  h/s.    \l  in. 
4').   A  girder  of  uniform  stmii^t/iy  of  length  /.  breadth  /'.  aii<l  d'  pili  ,/, 
rests  'jpo!    Mvo         lorts  and  carries  a  uniforndy  distril)nted  l^ad  oi  ,-,' Ihs. 
per  unit  1, :  le.,giii,  which  produces  an  inch-stress  of/  lbs.  at  eveiy  jioiiii 

n 


of  tlu 


';'rial.     Show  that  the  central   delleclion  is 


when  /' is  lOiisl.iMi  and  *,' \'iiriable.     Find  the  deflection  when  </  is  con- 
stant and  />  variable. 


yh/s. 


4  a;/' 


50.  A  seiui-girder  of  uitifonn  s/n-ii^i^t/i,  of  length  /,  breadlii  /-,  and 
depth  f/,  carries  a  weight  //'at  the  free  end  which  produces  an  inch- 
stress  of /"lbs.  at  every  point  of  the  material.     Prove  that  the  ma.ximiiin 

4   ( A/|3/    />   \» 

detlection  is ^r.-L  ,,%    wiien  /'  is  constant  and  ti  variable,  and  that 

3      A    \6/// 

it  is  twice  as  great  as  it  would  be  if  the  section  were  uniform  tliioii^hout 

and  e(]nal  to  that  at  the  supjiori. 

What  would    be   the    ma.xinumi    deflection    if  tlie  semi-girder  were 

subjected  to  a  uniformly  distributed  load  of  w  lbs.  per  unit  of  length .' 


.Ins. 


51.  The  neutral  a.xis  of  a  symmetrically  loaded  girder,  who.^e  .luiiiiciii 
of  inertia  is  constant,  assumes  the  form  of  an  elliptic  or  circular  arc. 
Show  that  the  bending  moment  at  any  point  of  the  deflected  girder  is 
inverselv  proportional  to  the  cube  of  the  vertical  distance  between  llu' 
point  and  the  centre  of  the  ellipse  or  circle. 

52.  A  vertical  row  of  water-tight  sheet  piling,  12  ft.  Iii!.;li,  is 
supported  by  a  series  of  uprights  placed  6  ft.  centre  to  centre  .ind 
securely  ti.\ed  at  the  base.  Find  the  greatest  deviation  of  an  ui)rii,'lit 
from  the  vertical  when  the  water  rises  to  the  iop  of  the  piling.  What 
will  the  maxiiiuim  deviation  be  when  the  water  is  6  ft.  from  the  top.-' 


Am. 


7.'/'//' 
30/:/ 


3 1 1 0400 


7i'0        ,  ^^  7('/>c-     ,  ^         21S72O 

LI         i'^EI  24A./  tl 


KXAMPIES. 


499 


,(•;• 

sni>|i<>rls 
is  1  2  in. 

1  Inad  oi 

1)S. 

2  ill.  wiiio 
:o,(H)u  llis. 
■.   \\  ill. 
:1  (IcpUi  ,/, 

,(l.)l7.'lllS. 

VflV  l'i>i"i 

/•;  \  ,v-''i 

I     ,/    IS    CDIl- 

<llh  /',  iiiid 

•cs  nil  incli- 

ina.Kiiiiiiiii 

0,  and  lluil 

ifoiij^hmit 

jirdor  were 
|f  lcm.;ili"' 

./■  /  *'■ 

l)>e  .iioiuciit 
lircular  arc. 
|il  irinler  is 
^'iwoi'ii  iIk' 

lii^li.   is 
Icentie  and 


,aii  upriii 


lit 


"K- 


What 


the  top : 


; 1 8720 


53.  A  vi'iiit:iil  ri)W  of  water  tiidit  sli'-ct  piling,  30  ft.  hi)^!',  is  siipiiorlcfl 
liv  a  scries  of  iipti^;iil.i  placed  .S  It.  c(!iilr(!  to  cetilrc  .infl  securely  (i.xed  at 
tin;  liasc,  while  the  upper  ends  are  kept  in  the  vertical  by  struts  slo[)inj3; 
ill 45  ■  If  the  water  ris<;s  t(j  the  to|)  of  the  pilitij^,  liiid  («)  the  thrust  on  a 
strut;  (b)\.\\M.  niaxiinuin  intensity  of  stress  in  an  iipri(.,dit;  (c)  the  amount 
and  position  of  the  inaxiuiiim  deviation  of  an  u|)ri}^lit  from  the  vertical. 

Alls.  45000^/2  lt)S.;  max.  H.  M.  —  -,  and  max.  intensity  of 

'54/5 

I    (  7f'//'        r    .S7.'//«     I  ^       . 

aticss  -^     ,  '     —  1  V —      ;    deflection    is  a  max.    when 

./  I     .0         /  ,5  ^'5    ) 


.r  = 


30 


,  and  lis  amount 


7.V/' 


.32 


4/5         \'% '■'  7S^Vs' 

v(  The  pilin;.,'  in  the  preccdinf^  example  is  strenj^thciicd  by  a  seconfl 
s.rics  of  struts  sloi)in).(  at  45"  from  tiie  points  of  maximum  deviation. 
Imiu!  the  normal  reactions  upon  an  upright  and  tiie  bcndinj;  moment  at 

I'S   llli  )l. 

What  will  be  the  reactions  and  bcndinj;  moment  if  the  second  row  of 
siiiiis  starts  from  the  middle  of  the  uprights? 

.his.  .007547.'//-';  .13771'/'';  .920277.'//»;      ,V()W/^';  ^JJw/^';  Ulw/i\ 

jy  A  continuous  girder  of  three  spans,  the  outside  spans  being 
wjual.  is  uniformly  loaded.  What  must  be  the  ratio  of  the  lengths  of 
ilif  centre  and  a  side  span  so  that  the  neutral  axis  maybe  hori/ontal 
over  t lie  intermediate  supjjorts.'  /^«5.    4^3  •  ^  ^■ 

56.  What  should  the  ratio  bo  if  the  centre  span  is  hinged  (n)  at  the 

centre;  (/;)  at  the  points  of  trisection  ?  /ifts. (a)  1^2  :  i  ;  (/;)  3  :  2  \^2. 

57.  Four  weights,  each  of  6  tons,  follow  each  other  at  fixed  distances 
of  5  ft.  over  a  continuous  girder  of  two  spans,  each  equal  to  50  ft.  If  the 
second  and  third  supports  are  i  in.  and  1^  in.,  respectively,  vertically 
iiclow  the  first  support,  find  the   maximum  H.  M.  at  the  intermediate 

sup[K)rt. 


Ans.  (.9855  -  ~ 


ft. -tons. 


40000/ 

58.  A  continuous  girder  of  two  equal  50-ft.  spans  is  fixed  at  one  of 
the  end  supports.  The  girder  carries  a  uniformly  distributed  load  of 
1000  lbs.  per  lineal  foot.  Find  the  reactions  and  bending  moments  at 
the  points  of  support.  How  much  must  the  intermediate  support  be 
lowered  so  that  it  may  bear  none  of  the  load  ?  How  much  should  the 
free  end  be  i/ten  lowered  to  bring  upon  the  supports  the  same  loads  as  at 
the  first  ? 

Ans.  Reactions-   !3,2[4f.  57,1425,  19,6425  lbs. ; 

Bending  moments  —  178,571!,  267,857}  ft.-lbs; 


coo 


T//EOUV  OF  SrKUCTU/^ES. 


m; 


\t  f  '> 


m^' 


59.  Four  loads,  each  of  12  tons  ami  spaced  5.  4,  and  5  ft.  apart,  travel 
in  oi'dcr  over  acotuinuousi^irdur  of  two  spans,  tlio  otinof  30  and  the  oilKr 
of  20  ft.  I'lace  the  wheels  so  as  to  throw  a  maximum  B.  M.  upon  the 
centre  support,  and  tind  the  correspondinii;  reactions. 

Draw  a  diagram  of  B.  M.,  and  find  the  maximum  deflection  of  each 
span. 

60.  The  loads  upon  the  wheels  of  a  truck,  locomotive,  and  tender, 
counting  in  order  from  the  front,  are  7,  7,  10,  10,  10,  10,  8,  8,  8,  8  tons. 
the  intervals  beinj,'  5,  5,  5,  5,  5,  9,  5.  4,  5  ft.  Tl  c  loads  travel  over  a 
continuous  girder  of  two  50-ft.  spans  AB,  DC.  Place  the  locomotive, 
etc.,  ((0  on  the  span  AB  so  as  to  give  a  maximum  B.  M.  at  B\  (/'i  so  as 
to  give  an  absolute  tiiaxiinum  B.  M.  at  B. 

61.  A  continuous  girder  of  two  spans /i^,  .flC  has  its  two  ends,/ 
and  C fixed  to  the  abutments.  The  load  upon  AB  is  a  weight  /'  disiaiu 
/  from  A,  an<l  that  upon  ^'Ca  weight  (2<J'**t^"t  q  from  C.  The  Icnyili 
of  ///)  =  /i,  of  />'C  =  /a.  The  bending  moments  at  A,  B,  C arc  J/i ,  M~. 
A/3,  respectively.  The  areas  of  the  bending-moment  curves  for  liie 
spans  AB,  BC  assumed  to  be  independent  girders  arcAi,  A-i,  respect- 
ively.    Show  that 

MJy  +  M^ih  +  /a)  +  M:,h  =  -  2{Ai  +  At), 

and  Mt{h  +  A)  =  —  2(^,/>  +  A^q). 

If  /,  =  /a  =  /,  show  that  il/a  is  a  tnaxiinuin  if 

62.  A  continuous  girder  of  two  spans  AB,  BC  rests  upon  supports  at 

A,  B.     A  uniformly  distributed  load  £"/-' travels  over  the  girder.    (/,  is 

the  centre  of  gravity  of  the  portion  BE  upon  AB,  and  6^3  iliat  of  the 

^ox\\o\\BF  upon  BC.     If  the  bending  moment  at  B  is  a  maxtiiiiiin.^hiiw 

that 

AE.EB      AG, 

CF .  FB  ~  CO', ' 

63.  An  eight-wheel  locomotive  travels  over  a  continuous  girder  (^f 
two  loo-ft.  spans ;  the  truck-wheels  are  6  ft.  centre  to  centre,  the  loaii 
upon  each  pair  being  8000  lbs. ;  the  driving-wheels  are  8;^  ft.  centie  to 
centre,  the  load  upon  each  pair  being  16.000  lbs.  ;  the  distance  centre  tu 
centre  between  ths  front  drivers  and  the  nearest  truck-wheels  is  alsn 
^\  ft.  Place  the  locomotive  so  as  to  throw  a  maximur"!  B.  M.  upon  the 
centre  support,  and  find  the  corresponding  reactions. 

64.  If  an  end  of  a  continuous  girder  of  any  number  of  spans  is  fixeH. 
show  that  the  relation  between  the  moment  of  fixture  (J/i)  and  the 

bending  moment  (il/a)  at  the  consecutive  support,  is  2.I/1  -f  J/a  =  — 


or  2;l/i  -I- J/a  =  —   .j2[ /'/(/  —  /)( 2/  —  /)],  according  as  the  load   iip! 


If  ■    ;  ! 


EXAMl'LES. 


501 


e  load   upon 


the  span  (/)  between  the  fixed  end  and  the  const  ciiiivc  support  is  of 
uniform  intensity  or  consists  of  a  number  of  weights /';.  Ft,  Pi,  .  .  . 
concentrated  at  points  distant  A,  A,  A,  .  .  ,  from  tho  lixcd  end. 

65.  A  continuous  girder  of  two  spans  AH,  BC,  carrying  a  load  of 
tuiilorm  intensity,  has  one  end  A  Jixfti,  and  the  other  end  rests  upon  the 
support  at  C.  If  the  bending  moments  at  A  and  B  are  equal,  show  that 
tiie  s|)ans  are  in  the  ratio  of  1^3  to  \^2,  and  find  the  reactions  at  the 
supports,  Wi  being  the  load  upon  ^IB,  and  W^a  that  upon  BC. 

Ans.   At  .-/  reaction  =  A  /f, . 

"  B      "      =  ur,  +  %ii\. 
"  c     "     =hv.. 

66.  A  viaduct  over  the  Garonne  at  Rordeaux  consists  of  seven  spans, 
viz.,  two  end  spans,  each  of  57.371;  ni.,  and  five  intermediate  spans,  each 
of  77.06  ni. ;  the  main  girders  are  continuous  from  end  to  end,  and  are 
t;icii  subjected  to  a  dead  load  of  3050  k.  per  lineal  metre.  Determine  the 
absolute  maximum  bending  moment  at  the  third  support  from  one  end. 
.\ls )  find  the  corresponding  reactions,  the  points  of  inflexion,  and  the 
liia.xinuim  deflection  in  the  first  and  second  spans. 

67.  A  continuous  girder  consists  of  two  spans,  each  50  ft.  in  length; 
the  effective  ilepth  of  the  girder  is  8  ft.  If  one  of  the  end  bearings 
settles  to  the  extent  of  i  in.,  find  the  maximum  increase  in  the  fiange 
and  siiearing  stress  caused  thereby,  and  show  by  a  diagram  the  change 
in  the  distribution  of  the  stresses  throughout  the  girder.  (.Assume  the 
section  of  the  girder  to  be  uniform,  and  take  /i  =  25,000,000  lbs.) 

Afis.  Increase  of  maximum  B.  M.  =  V/'W'    ^      —  i  )■ 

\2!67£'  / 

"        "  shearing  force       =  H^- 
70  being  weight  per  unit  of  length,  and  /the  moment  of  inertia. 

68.  A  girder  carrying  a  uniformly  distributed  load  is  continuous  over 
four  supports,  and  consists  of  a  centre  span  (/j)  and  two  equal  side 
spans  (A).  Find  the  ratio  of  A  to  A ,  so  that  the  neutral  axis  at  the 
intermediate  supports  may  be  horizontal.  Also  find  the  value  of  the 
laiiowhena  hinge  is  introduced  (a)  at  the  middle  point  of  the  centre 
span  ;  (/;)  at  the  points  of  trisection  of  the  centre  span  ;  {c)  at  the  middle 
points  of  the  half  lengths  of  the  centre  span, 

69.  In  a  certain  Howe  truss  bridge  of  eight  panels,  the  timber  cross- 
tie<  are  directly  supported  by  the  lower  chords,  and  are  placed  sufii- 
cientlv  close  to  distribute  the  load  in  an  approximately  uniform  inanner 
over  the  whole  length  of  these  chords,  thus  producing  an  additional 
stress  due  to  flexure.  Assuming  that  the  chords  may  be  regiarded  as 
girders  supported  at  the  ends  and  continuous  over  seven  intermediate 


1 


rnT-r-r- 

?1  t 


i^ 


I 

(i 


I 


Hm 


502 


TIfEOKY  OF  STRUCTURES. 


supports  coincident  with  the  pane!  points,  and  that  these  panel  puint- 
are  in  a  truly  horizontal  line,  determine  (a)  the  bending  moments  and 
reactions  at  the  panel  points  ;  {b)  the  maximum  intermediate  bending 
moments;  and  (c)  the  points  of  inflection,  corresponding  to  a  load  of  r^' 
per  unit  of  length,  /  being  the  length  of  a  panel. 

Ans. — (a)  At  ist  support ;  2d  support;  3d   support; 

B.  M.  =        o         ;    -sVf-f/';       -3V'.'''^^ 
Reaction  =  /?,  =  )i\\iul\  Ri  =  i\livl\  /?s  =  WVwl ; 

At  4th  sup[)(jrt ;  5th  support. 


B.  M.  =  -  aVVc//^ 


3Ss' 


Reaction  =  /^t  =  iJssw/;     /\\  =  ats''-'^- 
(fi)  Maximum  intermediate  B.  M   = 

6600.5 


1 1704.5    „. 

-—riiJi    in  1st  span; 

(388,1'  ^ 


5104.5     ,,    .         ,  6600.5     .„  . 

=  —Jc.  M     ill  2d  ;  =  —  ,:„-^w/'  n  3d  ; 
(388)-'  (388)'  -^ 


6208.5 
=  — T^TT-n'vl'^  in  4th. 

(<r)  Points  of    inflexion   in   the  four   spans   arc   given   bv 

2R1       1,06  w 

x= =  V^/  :  A',(/  4-  .1-)  +  ^.-i- (/  +  .vf  =  Q; 

w        3S8  2 

/?,(2/  4-  X)  +  A'a(/  4-  -t )  +  A'a.r  —  -(2/  +  .I)-  =  0; 
^,(3/  4-  X)  4-  Rii^l  4-  .1-)  4-  Aof/  4-  -r)  +  ^4.1"  —  -^Z  4-  -V)-  =  o. 

70.  A  continuous  girder  of  two  equal  spans  \^  fixed  at  one  of  the  end 
supports.  The  girder  carries  a  uniformly  distributed  load  of  intensity 
w.  If  the  length  of  each  span  is  /.  find  the  reactions  and  moment  of 
fixture.  How  much  must  the  intermediate  support  be  lowered  so  that 
it  may  bear  none  of  the  load  ?  How  much  should  the  free-end  support 
then  be  lowered  to  bring  upon  the  supports  the  same  loads  as  before } 


II        16       13    ,        luP 
Ans.  —wl,  — ivl,  -—ivl\  — 


5  wl"       5 


14 


14  '  48  £•/  •    24  /••/ 


71.  Each  of  the  main  girders  of  tlie  Torksey  Bridge  is  continuous 
and  consists  of  two  equal  spans,  each  130  ft.  long.  The  girders  are 
double-webbed  ;  the  thickness  of  each  web  plate  is  \  in.  at  the  centre 
and  f  in.  at  the  abutments  and  centre  pier;  the  total  depth  of  the  gir- 
ders is  10  ft.,  and  the  depth  from  centre  to  centre  of  the  flanges  is  9  ft. 
4I  in.  Find  (a)  the  reactions  at  the  supports,  and  also  {b)  the  points  of 
inflexion,  when  200  tons  of  live  load  cover  one  span,  the  total  dead  lo;i(! 
upon  each  span  being  180  tons  uniformly  distributed.     The  top  flange  is 


EXAMPLES. 


503 


mel  points 
menih  and 
te  bending 
I  load  of  -^ 

ipport ; 
Lipporl. 

in  I  si  span  ; 

in  3d ; 

in  4tii. 

•c   "ivcn   bv 


(/  +  -1-; 


:o; 


:/  +  .r)-  =  o; 

/  +  .V)'  =  o. 

of  the  cn'l 

|of  iiitensitv 

moment  of 

;red  so  that 

lend  support 

IS  before  ? 

24  y.i 

I  continuous 
(girders  aiv 

the  centn- 
of  tin-  i;ir- 
[nges  is  9  f- 
lie  points  i^f 

II  dead  load 
top  flange  is 


ceKulai  ;  \\.% gross  sectional  area  at  the  centre  of  each  span  is  51  sq  in  . 
and  the  corresponding  net  sectional  aiea  of  tiie  bottom  flange  is  55  sq. 
in.  Determine  (t)  the  flange  stresses  and  {U)  the  position  of  the  neutral 
axis.  (/  =  372,500.)  Also  {e)  determine  the  reactions  when,  first,  B  and, 
second,  6' are  lowered  i  in.  {E=  i6,yoo  tons.) 
Ans. — (a)  155.  350,  55  tons. 

(^b)  io6,'5  and  79J  ft.  from  end  support. 

(c )  6  7   and    7.3   tons  per  sq.  in.  in   loaded  span;    1.13   and 

1.22  tons  per  sq.  in.  in  unloaded  span. 

(d)  58.3  in.  from  centre  line  of  lop  flange. 


{e)  First,  /e,  =  155  + 


I  El 

4>' 


A',  =  350 


A'o  =  55  H- 


I  El 
Second,  i*?!  =  1 55  —  <.  ■^  :  A  =  35°  + 


R, 


55- 


I  EI 

'  2  P  ' 

I  EI 

4   I'' 
I  AY 

4  i^' 
I  EI 


8    /' 


£7 

/3 


Where 

18^5 
1 1 232 


72.  Two  tracks,  6  ft.  apart,  cross  the  Torksey  IJridge,  and  are  sup- 
ported by  single-webbed  plate  cross-girders  25  ft.  long  and  14  in.  deep. 
If  the  whole  of  the  weiglit  upon  a  pair  of  drivers,  viz.,  10  tons,  be  directly 
transmitted  to  one  of  these  cross-girders,  draw  the  corresponding  shear- 
ing force  and  bending-moment  diagrams  (i)  if  the  ends  of  the  cross- 
girder  are  yf.itv?' to  the  bottom  flanges  of  the  main  girders  ;  (2)  if  they 
merely  rest  on  the  said  flanges.  Find  the  iiiaxiimim  rleflection  of  the 
cross-girder  and  the  luorlc  done  in  bending  it,  in  each  case. 


Ans.  (I) 


7088.45 

-  at  13.208  ft.  from  one  end. 

_      ,         ,      ,  „  67161  6 

Total  work  of  flexure  =  .-    ft -tons. 


73.  A  swing-bridge  consists  of  the  tail  end  ///>',  and  of  a  span  BC  of 
length  I  ft.,  the  pivot  being  at  />.  The  balIast-bo.\  oi  weight  ff  extends 
over  a  length  AD  (=  zc  ft.),  and  the  weight  of  the  bridge  from  D  to  />' 
is  Ti' tons  per  lineal  foot.  If  DB  =  x.  if  /  is  the  cost  per  ton  of  tiie 
uridine,  and  if  g  is  the  cost  per  ton  of  the  ballast,  show  that  the  total  cost 


is  a  minimum   when   x  -f-  c 


_(g(C-c')\^ 


\  2/ 


,  and  that  the  corresponding 


weight  of  the  ballast  is  wxi—  —  i  J  4-  -wr. 

\9  I      Q 


.  ,_....,^s*.L..  M 


504 


THEORY  OF  STRUCTURES. 


74  Co:npa'C  graphically,  the  shcariiifj  forces  and  bending  moments 
along  the  span  BC  ot  the  bridge  in  the  preceding  question  when  the 
bridge  is  closed,  with  their  values  when  the  bridgf  is  open.  What  pro- 
vision should  be  made  to  meet  the  change  in  the  kind  oi  stress? 

75.  Each  of  the  main  girders  of  a  railway  bridge  resting  upon  two 
end  supports  and  five  intermediate  supports  is  fixed  at  the  centre  sup- 
port, is  3  ft.  deep  throughout,  and  is  designed  X.o  carry  a  uniformly  dis- 
tributed dcad\oi\A  of  \  ton  and  a  live  load  of  A  ton  per  lineal  foot.  The 
end  spans  are  each  51  ft.  8  in.  and  the  intermediate  spans  each  50  ft  in 
the  clear.  Find  the  reactions  at  the  supports.  'I'he  girders  are  sini^it;- 
webbed  and  double-flanged  ;  the  flanges  are  12  in.  wide  and  equal  ;n 
sectional  area,  the  areas  ff)r  the  intermediate  spans  i)eing  13  sq.  in.  luid 
17  sq.  in.  at  the  centre  and  piers  respectively.  Find  the  correspond  in;,' 
moments  of  resistance  and  flange  stresses,  the  web  being  f  in.  iliick. 

Ans.  Reaction  at  island  7th  supports  =  iSsonSill  '<  ^^  2d  and  iiii 

supports  =  43'^J5(iJ"  ;  ''^  3^  ^"^^  5^^  supports  =3;,';;;:; 

at  4th  support  =  38"///^*',  tons. 

Ax. piers— =  693  and  flange  stresses  are  3.59  tons  per  sq.  in. 


.at  2d  support,  2.45  at  3d,  and  2.£3  at  4th. 
At  Centre --=  549  and  flange  stresses  in  istspan 


3.-  tons 


per  sq.  in.,  in  2d  =  1.3,  and  in  3d  =  1.78. 
76.  A  continuous  beam  of  four  equal  spans  carries  a  uniformly  dis- 
tributed load  of  Ti'  intensity  per  unit  of  length.  The  second  support  is 
depressed  a  certain  distance  </ below  the  horizontal,  and  the  reaction  at 
the  2d  support  is  twice  that  at  the  ist.  Show  that  the  reactions  at  the 
1st,  2d,  3d,  4th,  and  5th  supports  are  in  the  ratio  of  the  numbers  15,  30, 
36.  34.  and  !  3 ;  find  d.  With  this  same  value  of  d  find  the  reactions  when 
one  end  isjixed. 

^'"•^'=4V^- 

TJ.  A  continuous  girder  of  two  equal  spans  (/)  is  uniformly  loaded. 
Show  that  the  ends  will  just  touch  their  supports  if  the  centre  support 


is  raised 


Sis/ 


78.  If  di ,  di ,  da ,  di  are  respectively  the  deflections  of  the  ist,  2d,  3d, 
and  4th  panel  points  in  question  69,  show  that  the  bending  moment  at 
the  middle  panel  point  {^f^)  is  given  by 

-93^/4  =  -  ^-~{6()d,  -  md,  +  24./,  -  6d,)  +  ii-7i'l\ 


/■..\',iMJ'/./:s. 


505 


moments 

when  the 

What  pro- 

upon  two 
enlrc  ^up- 
(ormly  (lis- 
foot.  'I'lie 
ch  50  fi.  in 

are  single- 
l  oqiial  in 
sq.  in.  and 
responding; 
I.  thick. 
I  2(1  and  jih 


IS  per  sq.  m. 

n  =  3.-  tons 

ormly  dis- 
supp(jrl  is 
reaction  at 
ions  at  the 
bers  1 5.  3*^' 
lotions  when 


-mly  loaded, 
itre  support 


ist,  2d,  3d, 
moment  at 


r. 


79.  .\  girder  supported  at  the  ends  is  30  ft.  in  the  clear  and   carries 

iw'i  stationary  loads,  viz.,  7  tcjiis  concentrated   at  6  ft.  and  (2  ions  at  18 

ft.  from  the  left  siipp(jrt.     Find  the  position  and  amount  of   the   maxi- 

nuiiu  deflection,  and  also  liie  work  of  flexure.     The  girder  is  built  up  of 

pl.itcs  and  an^U--iions  and  is  24  in.  deep.     If  the  moment  of  resistaiue 

line  to  the  Wfl)  is  ne),'lected,  and  if  the  iiUensiiy  of  the  longitudinal  stress 

IS  not  to  exceed  5  tons  per  sq.  in.,  wiiat  siiouid  be  the   flange  sectional 

area  correspond  in  v;  to  the  maxinuim  bending  moment. 

Ans.   Max   deflection  ^  i"ri' -  JCi' —  6)' —  ^V^-i'.  where 

.1=1  5.3+ ft. 

,„     ,        67i6r  6  , 
WorK  =  It.-tons. 

Sect,  area  =  10.32  sq.  in. 

So.  Determine  the  work  of  flexure  and  tin-  necessary  flange  sectional 
area  at  tlie  ci'iuic  if  tlie  i;irder  in  the  preceding  question  is  subjected 
to  a  uniformly  distributed  load  of  40  t(jns,  instead  of  the  isolated  loads. 


Alls.  Work 


540000. 

'    tt.-tons;  sect,  area  =  15  sq.  in. 


Si.  {(I)  The  bridge  over  the  Garoime  at  Langon  carries  a  double 
track,  is  about  695  ft.  in  length,  and  ci>nsistsof  three  spans,  .//>', />C,  fZ?. 
The  two  main  girders  are  continuous  and  rest  upon  the  abutments  at 
./  and  D  and  upon  piers  at  />'  and  C.  The  effective  length  of  each  of 
the  spans  .IB,  CD  is  20S  ft.  6  in.,  and  of  the  centre  span  y)'C'243  ft.  The 
permanent  load  upon  a  main  girder  is  1277  lbs.  per  lineal  foot,  and  the 
proof  loafl  is  2688  lbs.  per  lineal  foot.  Find  the  reactions  at  the  sup- 
ports (I)  when  the  pr<j(jf  load  covers  the  span  .//>';  (2)  when  the  proof 
load  covers  the  span  BC ;  (3)  when  the  proof  load  cover  the  spans  AB 
and  liC;  (4)  when  tlie  proof  load  covers  the  whole  girder. 

Draw  shearing-force  and  bending-moment  diagrams  for  each  case. 

(b)  At  the  piers  i.ie  web  is  i  in.  thick  and  18  ft.  in  depth,  and  each 
flange  is  made  up  of  four  plates  A  in.  thick  and  3  ft.  wide.  Determine 
the  flange  stresses  for  cases  (i  j  and  (3). 

1(1  The  angle-irons  connecting  the  flanges  with  the  web  at  the  pier 
are  riveted  to  the  former  with  (J-in.  rivets  and  to  the  latter  with  i-in. 
rivets.  How  many  of  each  kind  are  required  in  one  line  per  lineal  foot 
on  both  sides  of  the  pier  at  B,  Sooo  lbs.  per  square  inch  being  safe 
shearing  stress? 

(d)  The  effective  height  of  the  pier  at  B  is  41  ft.,  its  mean  thickness 
is  14  ft.  9  in.,  its  width  is  42  ft.  9  in.,  and  it  weighs  125  lbs.  per  cubic 
foot.  If  there  is  no  surcharge  on  the  bridge,  and  if  the  coefficient  of 
friction  between  the  sliding  surfaces  at  the  top  of  the  pier  is  taken  at 
.15,  show  that  the  overturning  moment  due  to  the  dilatation  of  the 
girders  is  about  y'j  of  the  amount  of  stability  of  the  pier. 


5o6 


THEORY  OF  STRUCTURES. 


«  ,T  *■  5    • 


i<       «      tt 

<t       <«      It 


<(         <(       II 


(c')  Find  the/t^/Vj/j  0/  inflexion  and  also  the  maximum  deflections  in 
Case  3. 

What  practical  advantage  is  derived  from    the  calculation  of  the 
deflection  ? 

An5.—{a)   Case  I.  /i'l  =  353469.95;  ^g  =  656955.7 ; 

Jit  —  2S0612.55  ;  /v'l  =  10900S.77  lbs. ; 
iJ/3= —1 22471 15.3:  iJ/3  =— 4823424.5  ft.-ibs. 
Case  2.  A'.  =  681 85  2  =  Rt  lbs.  ; 
A'.j  =  6791783  =  /v's  lbs.; 
,1/5=  —  13439537.7  ft.-lbs.  =  M,. 
Case  3.  A',  =  312982.65  ;  J\\  =  1024035 ; 

A'3  =  647691  ;        A'4  =69121.47  lbs.  ; 
iJ/3=  —  1 565031.2  ;  A/a  =  —  8226621.2  ft.-lbs. 
Case  4.  /?!=  4-2591.42    =i?4lbs.; 
A'a  =  1304647.  55  =  A3  lbs. ; 
i1/3=  —  15455566  =  Ma  ft.-lbs. 
(^   /«  2130816;  in  case  i./a  =  7448.9  lbs.  persq.  in. 

.   /«  =  2933-6    " 
in  case  3, /a  =  9400.3    " 
73  =  5003.5 

(Weakening  effect  of  rivet-holes  in  tension  flange  is 
neglected.) 
(f)    9.1  per  lineal  foot;  11.5  per  lineal  foot. 

(d)  Moment  of  stability   =  23833291 'js|  ft.-lbs.  ; 
overturning  moment  =  1919408.S  ft.-lbs.; 

ratio  =  12.4. 

(e)  Points  of  inflexion:   in  AH,  157.8  ft.  from  A;   in  BC, 

at  a  distance  x  from  B  given  by  .i"—25Sj.i- -1-104261 
=  0;  in  CD,  at  54.1  ft.  from  D. 
Max.  deflections . 

In  AB.        -ttX^^Sti-^'*  —  52163.7.1'  4-  227693091.6.1), 
A/ 

where  .r  is  given  by  66oi!.r-—  1 56491 .3.1+  227693091.6=0  ; 

In  BC,  -^-(i65.2.i-*-853829.i-' 4-9977485. 9.1'^- 103272869681, 
A/ 

where  .v  is  given  by  .v'  —  3876.r  -j-  30196^  =  o. 
82.  A  beam  AB  oi  span  /  carrying  a  uniformly  liistributed  loail  of 
intensity  w  is  fi.xed  at  A  and  merely  supported  at  B.     The  end  />'  is 

lowered  by  an  amount  —r-r.-     Find  the  reactions.     How  mucii  nui.st  I> 

16A/ 

be  lowered  so  that  the  whole  of  the  weight  may  be  borne  at  .-/? 

I  lel* 


Ans.  {\iifl  at  A,  {^wl  at  B  ;  --rr.- 


EXAMPLES. 


507 


83.  Solve  the  preceding  question  supposing  the  fixture  at  A  to  be 
imperfect,  the  neutral  axis  making  with  the  horizontal  an  angle  whose 
I  wl* 


7  wl^ 


tangent  is  — „  -^.  Ans. 

84.  A  wrought-iron  girder  of  I-section,  2  ft.  deep,  with  flanges  of  equal 
area  and  having  their  joint  area  equal  to  that  of  the  web,  viz.,  48  sq.  in., 
carries  \  ton  per  lineal  foot,  is  loo  ft.  long,  consists  of  five  equal  spans, 
and  is  continuous  over  six  supports.  Find  the  reactions  when  the  third 
support  is  lowered  \  in.  How  much  must  this  support  be  lowered  so 
that  the  reaction  may  be  nil  at  (<?)  the  ist  support ;  (/')  the  3d  ;  (c)  the 
jth.'  'How  much  must  the  support  be  raised  so  that  the  reaction  mav 
be  «//at  {(i)  the  2d  ;  {e)  the  4th  ;  and  (/)  the  6Lh  support.'  E—  16,500 
tons. 

Ans.  /?,  =  22| ;     /?=  =  I  ^l\  ;  A  =  3^^ ; 

A'4  =  14*;,;  ;  Rt  =  9i"j  ;  lit  =  4,^5  tons. 
(a)  i§  in.;  {/>)  Ul  '"■:  W  2 j^  in. ; 
(d)  i/3»rin.;  (e)  2,,f^in.:  (/)  6i  in. 

85.  If  the  three  supports  of  any  two  equal  ccjnsecutive  spans  of  a 
continuous  girder  of  any  number  of  spans  are  depressed  below  the 
horizontal,  show  that  tiie  relation  between  the  three  bending  moments 
at  the  supports  will  be  unafTected  if  the  depression  of  the  centre  support 
is  a  mean  between  the  depressions  of  the  other  two  supports. 

86.  A  girder  consists  of  two  spans  AB,  BC,  each  of  length  /,  and  is 
continuous  over  a  centre  pier  B.  A  uniform  load  of  length  2a{<  / )  and 
of  intensity  w  travels  over  ^/i')'.  Find  the  reactions  at  tiie  supports  for 
any  given  position  of  the  load,  and  show  that  the  bending  moment  at 

,  ...  ,  ,  (I'^l  I         '''  '      , 

the  centre  pifir  is  a  maximum  and  equal    to         -[i  —  -jr,\    when   the 

3  V'3  V         ^7 

87.  A  continuous  girder  rests  upon  three  supports  and  consists  of 
two  unequal  spans  AB  (=  /,),  BC  (=  A).  A  uniform  load  of  intensity  w 
travels  over  AB,  and  at  a  given  instant  covers  a  length  .ID  (=  r)  of  the 
span.     \l  Ri,  Ri  are  the  reactions  at  .1  and  C.  respectively,  show  that 

Rxh'  +  RJ^'  =  wrU'  -  K-/,  +  ^^ 

Draw  a  diagram  showing  the  shearing  force  in  front  of  the  nuwing 
load  as  it  crosses  the  girder. 

88.  If  the  live  load  in  the  preceding  question  may  cover  both  spans, 
show  that  the  shearing  force  at  any  point  Z*  is  a  maximum  when  .ID 
and  BC  are  loaded  and  BD  unloaded. 

Illustrate  this  iatze  graphically,  taking  into  account  the  dead  load 
upon  the  girder. 


centre  of  the  load  is  at  a  distance 


3  V'3 
from  /i. 


11 


^^%.4|^;5jj 


I 


Vn 


.1 

■n  i 


508 


THEORY  OF  STRUCTUKES. 


89.  A  continuous-girder  bridge  has  a  centre  span  of  300  ft.  and  two 

side  spans,  each  of  200  ft.     The  dead  load  upon  each  of  the  main  girders 

is  1250  lbs.  per  lineal  foot.     In  one  of  the  side  spans  there  is  also  an 

additional  load  of  2500  lbs.  per  lineal  foot  upon  each  girder.     Finfl  ilie 

reactions  and  points  of   inflexion.     How  much  must  the  third  suppcjii 

from  the  loaded  end  be  lowered  so  that  the  pressure  upon  it  may  be  jusi 

zero } 

Ans.  Let  IV  =  weight  on  loaded  span  =  750,000  lbs. 

i^.  =  iVA  ^  lbs. ;  A',  =  i  §f  I IV  lbs.; 

^.  =  iVW  ^^  lbs. ;  A'.  =  iVs'e  ^^  lbs. 

Af,  =  -  i!,V/  ^y  ft. -lbs.;  A/»  =  -  V/fi'  ^^  ft.     .s. 
Distance  of  point  of  inflexion  in  loaded  span  from  nearest 

end  support  =  i62i5','  ft. 
Distance  of  point  of  inflexion  in  unloaded  end  span  from 

nearest  end  support  =  145^?  ft. 
Distance  of  point  of  inflexion   in  intermediate  span  from 

end  support  in  unloaded  span   is  the  value  of  .r  in  the 

equation  .r*  —  '|','".i-  +  ^5^5"^^  =  o. 

,     ,  ,       ,.  56350000   fF 

3d  support  must  be  lowered  a  distance  = —. 

87        E/ 

90.  A  continuous  girder  AC  consists  of  two  equal  spans  AB,  BC,  eacii 
of  length  /,  and  carries  a  uniformly  distributed  load  of  intensity  icn  upon 
AB,  and  of  intensity  7£'a  upon  BC.  Determine  the  bending  moments  at 
the  supports,  the  maximum  intermediate  bending  moments,  and  the  re- 
actions {a)  when  both  ends  of  the  girder  are  fixed;  {b)  when  one  end  ./ 
is  fixed  and  the  other  free. 

Ans.  Denoting  tiie  reactions  and  bending  moments  at  A,  B,  C 

by  Bi ,  Af,  ,  A'a ,  Afi ,  A's ,  Ma ,  respectively  : 

/'  /' 

{a)  Afi  =  --  (—  574y,  +  Wa) ".    ^J/a  = (wi  +  Wj); 

48  24 

/"  /?  ' 

Ml  =-i(wi  —  5Wa);  Af,„ax.  in  AB  =  -—  +  Afi ,  in  BC 

48  27l>i 


=  —  +  Af, 


2Wi 


^'-76^- 


/?,  =  -(9W1 

10 


Wt);   Bi=  -(wj+wj); 


■7V:  +  gWa). 


(6)  Mi  = -(30/1  —  Wa);  Afi  = 

28 


28 


{wi  +  2Wi); 


Aft  =  o ;  ATmax.  in  AB  =  —  -V  Afx,    in  BC  =  — ; 

27£/i  2W, 

/  / 

/?,  =  -—(16a/,  —  3Wa);    A',  =  -r-(l3Wi  +  l9Wa); 
28  28 

/?,  =  —A—W,   +    I27£/a). 
20 


EXAMPLES. 


509 


91.  In  the  preceding  question,  if  lOx  =  Wq  =  «/,  find  the  points  of  in- 
flexion and  tlie  maximum  deflection  in  each  case  and  for  each  span. 

Ahs. — (a)  Points   of    inflexion    for   ////  or  BC   are   given   by 

e.v"  -  6.1V  +  /»  =  o. 
Max.  deflection  for  AJ>'  or  J>'C  is  given  by 

,        -E/y= (a/.v-.r'-/"). 

24 

in  wliich  the  value  of  .v  is  found  from 


!.r'  -  3/.r  +  /  = 


o. 


(d)  Points  of  inflexion  in  AB  are  given  by 

14.1"  —  i3.r/  +  2/"  =  o,   and  ;n  BC  by  .r  =  ^^/, 
Max.  deflection  for  ^IB  is  given  ijy 


-y^/v. 


7t7.t-' 


168 


,  (i3/.r- 6/' -7.r'), 


and 


28a-'  —  39/.r  +  12/"  =  o. 
Max.  deflection  for  BC  is  given  by 


4/'), 


and 


28^'  -33-1-'  4-  4/'  =  o. 
92.  A  continuous  girder  AC  consists  of  two  equal  spans  AB,  BC  of 
15  m.  each.  Determine  the  bending  moments  at  the  supports,  the  maxi- 
mum intermediate  bending  moments,  and  the  reactions  (a)  wiien  the 
load  upon  each  span  is  3000  it.  per  metre ;  {b)  wh(;n  the  load  per  metre  is 
3000  iv.  upon  y//j' and  1000  1<.  upon  BC.  Call  J/i  ,  .I/a,  J/s  the  bending 
moments  and  Ri ,  A'a ,  A'a  the  reactions  at  A,  /i,  C,  respectively,  and  con- 
sider three  cases,  viz.,  when  both  ends  of  the  girder  arc  free,  when  both 
ends  are  fixed,  and  when  one  end  is  free  and  the  other  fixed. 
Alts.  —Case  I : 

(a)  Ml  =0  =  Mi ;  Mi  =  —  84375  km.  ;  M„,ax.  in  AB  or  BC 
=  47460.9375  km. 
/?,  =  A'a  =  16875X' ;  A\  =  56250/(-. 
{b)  Mt  —  o  =  J/a  ;    Afj   ~  —  56250  km.  ;    Mmax.  in  AB 
=  58593.75  km.,  in  /)'t' =  7031.25  km. 
/?,  =  18750  k.  ;  A'a  =  37500  k.;  A\  —  3750  k. 


Case  II; 


(a)  M:  =  M,  =  M,  ■■ 
=  28125  km. 

A" 


—  56250  km.;  Mmax.  in  AB  or  BC 


/?,  =    -  =  A'a  =  22500  k. 


III 


H 


i 


M 


"I;' 


h  I 


510 


THEORY  OF  STRUCTURES. 


'n     •  {b)  M\  =  —  65625  km. ;   Af,  =  —  37500  km.  \  Ms  =  —  9375 

km.     Mmax.    in  .4//  =  33398.4375   km.,  in   BC 
=  6445.3125  km.; 
Ri   =  24375  k.;  A'a  =  30000  k.;  R3  =  5625  k. 
Case  III : 

{a)  Ml  =  —  48214?  km.;  JAj  =  —  72321^  km.;  J/:,  =  o. 
M,„a.v.  ill  An  -—  24537^5^  km.,  in  BC  =  52088}^'^  l<m.; 
R,   =  20892 J  k.;  A',  =  51428^  k.;  A'3  =  4821^'  k. 
{b)  yJ/i  =  -  64285^'  km.;  M;  =  -  40178^  km.;  J/a  =  o. 
Mm<t.v.  ill  ^^' =  — 32573 iVff  km.,    in    BC  =  wGzi^^^,^ 

km. ; 
Ri  =  24107I  k.;  i?a  =  3io7iJ  k.,  A'.  =482[^  k. 

93.  Show  that  a  uniformly  loaded  and  continuous  girder  of  two  equal 
spans,  with  both  ends  fixed,  is  2.08  times  as  stiff  as  if  the  ends  were  Irce 
and  merely  rested  on  the  supports. 

94.  A  single  weight  travels  over  the  span  AB  of  a  girder  of  two  equal 
spans,  AB,  BC,  continuous  over  a  centre  pier  B.     Show  that  the  reaction 

AB 
at  C  is  a  maximum  when  the  distance  of  the  weight  from  A  is     y_  il  the 

V3 
ends  A  and  C  rest  upon  their  supports,  and  when  the  distance  is  ^.//>'  if 
the  two  ends  are  fixed.     Find  the  corresponding  bending  moments  at 
the  central  pier. 

PI     2 
Ans.  ---;  -PI. 
6V3   =7 

95.  A  girder  with  both  ends  fixed  carries  two  equal  loads  W^'at  points 
dividing  the  girder  into  segments  a,  b,  c.  Determin'e  the  reactions  and 
bending  moments  at  the  supports. 


Ans, 


2rt'  +  (iii'b  4-  yib"^  4-  <^'  +  6rtV  4-  dabc  +  3<^V 


R-,  =  IV 

Ml  =  IV 
M-,  =  W- 


{a  +  b  +  o* 
2rtV  4-  2abc  4-  be''  4-  ab\ 
{a  +  b  +  c-y         ' 
lac'  4-  2abc  4-  (I'b  4-  b'^c 


(<i  4.  ^  4.  cf 

96.  A  bridge  a  ft.  in  the  clear  is  formed  of  two  cantilevers  wliicii 
nuet  in  the  centre  of  the  span  and  are  connected  by  a  bolt  capable  of 
transmitting  a  vertical  pressure  from  the  one  to  the  other.  A  weii;iit 
W  is  placed  at  a  distance  b  from  one  of  the  abutments.  Find  the  pres- 
sure transmitted  from  one  cantilever  to  the  other,  and  draw  the  curve 
of  bendi'  9;  moments  for  the  loaded  cantilever. 


Ans.    Ri=  lV[i  -3  .  +  2 


EXAMPLES. 

1 

■■     .■■■'.:■                ■    ■  ■  V  •■" 

5" 

97 

.  The  weights  7,  7,  10.  10,  10,  10,  8,  8,  8,  8  tons, 

taken  in  order  pass 

over 

d  continuous  g 

rder  of  two  spans,  each  of  50  ft. 

and  fixed  at  both 

ends, 

the  successive 

intervals  being  5,  5,  5,  5,  5,  9,  5 

,  d,  5  ft.     Place  the 

wheels  so  as  to  give 

the  maximum  bending  moment 

at  the  centre  sup- 

port, 

and  find  its  val 

ue. 

Ans.  First  wheel  25.8399  ft.  from  nearest  abutment ; 

Max.  B.  M.  =  306.62  ft.-tons. 

'    ]r  \ 


98.  The  bridge  over  the  Grande  Baise  consists  of  two  equal  spans  of 
19.8  in.;  each  ot  the  m^in  girders  is  continuous  and  rests  upon  abut- 
inciiis  at  the  ends.  Find  the  position  of  the  points  of  inflexion,  the 
bending  moment  at  the  centre  support,  the  maximum  intermediate 
b  luling  moment,  and  the  maximum  flange  stress  {ti)  under  the  dead 
l(,ad  of  1700  k.  per  lineal  metre;  {/>)  under  the  same  dead  load  together 
with  an  additional  prcjof  load  of  2000  k.  per  lineal  metre  on  one  span. 
Tlie  depth  of  the  girder  =  3.228  m.,  and  /=  .093929232444. 

Ans. — (ti)   14.85  m.from  the  abutments  ;  S3308.5  kilogrammetres 
(km.);  46,8613'j  km.;  1.4315  k.persq.  mm. 
{/>)   16.18    m.  from  abutment   on  loaded  side;    11.876  m. 
from    abutment    on    unloaded    side;.  132313.5  km.; 
101991. 65625  km.;  2.27356  k.  per  sq.  mm. 

99.  The  Estressol  viaduct  consists  of  four  spans  of  25  m.  ;  the  main 
girders  are  continuous  and  tiieir  ends  rest  upon  abutments;  the  dead 
load  upon  each  girder  is  1700  k.  per  lineal  metre.  Determine  the 
position  of  points  of  inflexion  in  each  span,  the  reactions  and  bending 
moments  at  the  supports  when  an  additional  load  of  2000  k.  per  lineal 
metre  crosses  (n)  the  ist  span;  (/^)  the  1st  and  2d  spans;  (c)  all  the 
spans. 

Also,  find  the  absolute  maximum  bending  moments  at  the  inter- 
mediate supports. 

Ans,  Call  .ti ,  Xi  .va,  .w  the  distances  of  points  of  inflexion  in 
1st,  2d,  3d,  and  4th  spans  from  the  ist,  2d,  4th,  and 
5th  supports,  respectively;  R\,  R-i,  A'a,  R^,  Rt,  the 
reactions;  J/i  ('=0).  J/a,  Mt,  Ah,  J/s  (=0)  the 
bending  moments. 

(a)  ,r,  =  20.72  m.  ; 

xj  is  given  by  1700X2"  —  4240174x3  +  395089^  =  o; 
x%  by  1700.1-3"  —  47767f.r.i  4-  238839^  =0 ;  .1-4  =  19.38  m. 
Rx  =  38348V'4  k-;  ^i  =  8n6of  k. ;  /?»  =  34107+  k.; 
Rt  =  49910?^  k. ;  Rt  =  i6473y\  ^■ 
M.,=  i97544i-^~  km. ;  Ah  =  535717  km. ; 
Mt=  4776^4  km. 


;  1  :     i 


mm 


lid 


U-,H 


11   ! 

II   I 


512 


THEORY  OF  STRUCTURES. 


(d)  .1-1  =  19.556  m. ;  3700.1,'  —  losooo.ra  +  503571!,'  -  o; 

I700.r3'''  —  41071^.13  4-  205357I  =  o;  .r.  =  20. 1(6  111. 

A',  =  36178T  k. ;  A\  =  107821^^  k.;  A'3  =  629645  k.; 

A'*  =  45^92?  k.;  A\  =  17142!  k. 

Mi=  258928f  km. ;  M3  =  120535^  km.; 

Mt—  102678^  km. 
(fi)  xi  =  19.64  m.  ;  .t'a  and  .vj  are  given  by 
i4-i'-375-'  + i«75  =  o. 

J/j=  A/i  =  247767^  km.  ;  J/3  =  165178;}  km. 
Abs.  max.  B.  M.  at  2(1  support  (=  max.  H.  M.  at  4th  sup- 
port) occurs  when  1st,  2(J,  and  4tli  spans  are  loaded, 

and  —  264508}!  km. 
Abs.  max.  B.  M.  at  3d   support  occurs  when  2d   and  3d 

spans  are  loaded  and  =  2098211'  Urn. 

100.  In  the  preceding  question  find  the  iibsohite  maximum  Hani^'c 
unit  stress  at  the  piers,  /  being  .093929232444.     ^/is.  4.5  k.  per  sq.  mm. 

101.  The  Osse  iron  viaduct  consists  of  seven  spans,  viz.,  two  ciui 
spans  of  28.8  m.  and  five  intermediate  spans  of  38  m.  ;  each  main  girder 
is  continuous  and  carries  a  dead  load  of  1450  k.  per  lineal  metre.  Find 
the  bending  moments  at  the  supports  when  a  proof  load  of  2250  k.  per 
line;d  metre  for  each  girder  covers  all  the  spans;  and  also  find  the 
absolute  maximum  bending  moment  at  the  fourth  support.  Is  the 
following  section  of  sufficient  strength  ? — two  equal  flanges,  each  com- 
posed of  a  6oo-mm.  x  8-mm.  plate  riveted  by  means  of  two  loo-nim. 
X  loo-mm.  X  12-mm.  angles  to  a  6oo-mm.  x  lo-mm.  vertical  web  plate 

and  two  80-mm.  x  8o-mm.  x  ii-mm.  angles  riveted  to  each  horizontal 
plate  with  the  ends  of  the  horizontal  arms  15  mm.  from  the  edges  of 
the  plates;  the  whole  depth  of  the  section  being  4.016  m.,  and  the  dis- 
tance between  the  web  plates,  which  is  open,  being  2.8  m.  If  insuf- 
ficient, how  would  you  strengthen  it.-" 

Ans.  J/a  =  416,518  km.;  J/s  =  452,790  km.;  Mk  =  443.722  km. 
Max.  B.  M.  =  542,199  km.     /  =  .14074440467. 


.•.  —  =  .07009183, 
and  max.  fiance  stress  =  -^  =  7.73  k.  per  sq.  mm. 


I 


This  is  much  too  large.  The  section  maybe  strengthened 
by  adding  two  6oo-mm.  x  8-mm.  plates  to  each  flan.i;e. 
/  is  thus  increased  by  .0783425536,  and  the  flange  unit 
stress  becomes  5  k.  per  sq.  mm. 


':1P 


CHAPTER  VIII. 


PILLARS. 


1.  Classification. — The  manner  in  which  a  material  fails 
under  pressure  depends  not  merely  upon  its  nature  but  also 
upon  its  dimensions  and  form.  A  short  pillar,  e.g.,  a  cubical 
block,  will  bear  a  weight  that  will  almost  crush  it  into  powder, 
while  a  thin  plank  or  a  metal  coin  subjected  to  enormous  com- 
pression will  be  only  condensed  thereby.  In  designing  struts 
or  posts  for  bridges  and  other  structures,  it  must  be  borne  in 
mind  that  such  members  have  to  resist  buckling  and  bending  in 
addition  to  a  direct  pressure,  and  that  the  tendency  to  buckle 
or  bend  increases  with  the  ratio  of  the  length  of  a  pillar  to  its 
least  transverse  dimension. 

Hodgkinson,  guided  by  the  results  of  his  experiments, 
divided  all  pillars  with  truly  flat  and  firmly  bedded  ends  into 
three  classes,  viz. : 

(A)  Short  Pillars,  of  which  the  ratio  of  the  length  to  the 
diameter  is  less  than  4  or  5 ;  these  fail  under  a  direct  pressure. 

(B)  Medium  Pillars,  of  which  the  ratio  of  the  length  to  the 
diameter  exceeds  5.  and  is  less  than  30  if  of  cast-iron  or  tim- 
ber, and  less  than  60  if  of  wrought-iron  ;  these  fail  partly  by 
crushing  and  partly  by  flexure. 

(C)  Long  Pillars,  of  which  the  ratio  of  the  length  to  the 
diameter  exceeds  30  if  of  cast-iron  or  timber,  and  60  if  of 
\vroui,fht-iron  ;  these  fail  wholly  by  flexure. 

2.  Further  Deductions  from  Hodgkinson's  Experi- 
ments.— A  pillar  with  both  ends  rough  from  the  foundry  so 
that  a  load  can  be  applied  only  at  a  few  isolated  points,  and  a 
pillar  with  a  rounded  end  so  that  the  load  can  be  applied  only 

•■'■"-/  ,    ..    --.::-  •    ■'        ■■.■      513 


n:  1 


{  i  ^ 


514 


THKORY  OF  STRUCTURES. 


along  the  axis,  are  each  one-third  of  the  strength  of  a  pillar  of 
class  B,  and  from  one-third  to  two-thirds  of  the  strength  of  a 
pillar  of  class  C,  the  pillars  being  of  the  same  dimensions. 

The  strength  of  a  pillar  with  one  end  flat  and  the  other 
round  is  an  arithmetical  mean  between  the  strengths  of  two 
pillars  of  the  same  dimensions,  the  one  having  both  ends  flat 
and  the  other  both  ends  round. 

Disks  at  the  ends  of  pillars  only  slightly  increase  their 
strength,  but  facilitate  the  formation  of  connections. 

An  enlargement  of  the  middle  section  of  a  pillar  sometimes 
increases  its  strength  in  a  small  degree,  as  in  the  case  of  solid 
cast-iron  pillars  with  rounded  ends  which  are  made  strongei- 
by  about  onc-scventh ;  hollow  cast-iron  pillars  are  not  affected. 
The  strength  of  a  disk-ended  pillar  is  increased  by  about  one- 
cis;hth  or  one-ninth  when  the  middle  diameter  is  lengthened  by 
50  per  cent.,  but  for  slight  enlargements  the  increase  is  imper- 
ceptible. 

The  strength  of  hollow  cast-iron  pillars  is  not  affected  by  a 
slight  variation  in  the  thickness  of  the  metal,  as  a  thin  shell  is 
much  harder  than  a  thick  one.  The  excess  above  or  deficiency 
below  the  average  thickness  should  not  exceed  25  per  cent. 

3.  Form. — According  to  Hodgkinson,  the  relative  strengths 
of  long  cast-iron  pillars  of  equal  weight  and  length  may  be 
tabulated  as  follows : 

(rt)  Pillars  with  flat  ends. 

The  strength  of  a  solid  round  pillar  being  lOO, 
"  "         "     square      "         is        93; 

*'  "         *'  triangular  "         is       1 10. 

(/;)  Pillars,  with  round  ends,  i.e.,  ends  for  hinging  or  pin 
connections. 

The  strength  of  a  hollow  cylindrical  pillar  being  100. 
•^  '•  "  "  an  H -shaped  "         is     74.6: 

*'  "  a  4-shaped  "         is     44.:: 

The  strengths  of  a  long  solid  round  pillar  with  flat  ends, 
and  a  long  hollow  cylindrical  pillar  v/ith  round  ends,  are  ap- 
proximately in  the  ratio  of  2.3  to  I. 

The  stiffcst  kind  of  wrought-iron  strut  is  a  built  tube,  the 


•1" :  1  ■>?'" 


.  pillar  of 
igth  of  ;i 
ions. 

the  other 
IS  of  two 
1  ends  flat 


ease 


their 


sometimes 
,se  of  suVul 
le  stronger 
ot  affected. 
^'  about  ("'('• 
igthened  by 
^se  is  imper- 

iffccted  by  a 
thin  shell  is 
1-  deficiency 
per  cent. 
ve  strengths 
gth  may  be 


lOO, 

93-. 
no. 
Iging  or  pin 

|r  being  lOO. 
is  lA-^'- 
is     44'-- 

Ith  flat  eivis, 
[nds,  are  ap- 


THE   FAILURE   OF  PILLARS. 


515 


m 


Fig.  335. 


section  consisting  of  a  cell  or  of  cells,  which  may  be  circular, 
rectangular,  triangular,  or  of  any  convenient  form. 

In  experimenting  upon  hollow  tubes,  Hodgkinson  found 
that,  other  conditions  remaining  the  same,  the  circular  was  the 
strongest,  and  was  followed  in  order  of  strength  by  the  square 
xwfour  compartments  i+1 ;  the  rectangle  \x\  /wt?  compartments, 
rr  I :  the  rectangle,  o  ;  and  the  square. 

The  addition  of  a  diaphragm  across  the  middle  of  the  rect- 
angle doubled  its  resistance  to  crippling. 

4.  Modes  of  Failure. — The  manner  in  which  the  crush- 
ing of  short  pillars  takes  place  depends  upon  the  material,  and 
the  failure  may  be  due  to  splitting,  bulging,  or  buckling. 

(a)  Splitting  into  fragments  is  characteristic  of  such  crys- 
talline, fibrous,  or  granular  substances  as  glass,  timber, 
stone,  brick,  and  cast-iron. 

The  compressive  strength  of  these  substances  is 
much  greater  than  their  tensile  strength,  and  when  they 
fail  they  do  so  suddenly. 

A   hard  vitreous   material,    e.g.,    glass   or   vitrified 
brick,  splits  into  a  number  of  prisms  (Fig.  335). 

A  fibrous  material,  e.g.,  timber,  and  granular  materials,  e.g., 
cast-iron  and  many  kinds  of  stone  and 
brick,  shear  or  slide  along  planes  oblique 
to  the  direction  of  the  thrust,  and  form 
one  or  more  wedges  or  pyramids  (Figs. 

336,  337,  338). 

Sometimes  a  granular  or  a  crystalline  substance  will  sud- 
denly give  way  and  be  reduced  to  powder. 

(/')  Bulging,  i.e.,  a  lateral  spreading  out,  is  characteristic  of 
blocks  of  fibrous  materials,  e.g.,  wrought  iron,  copper,  lead,  and 
timber,  and  fracture  occurs  in  the  form  of  longitudinal  cracks. 

All  substances,  however,  even  the  most  crystalline,  will 
bulge  slightly  before  they  fail,  if  they  possess  some  degree  of 
toughness. 

(()  Buckling  is  characteristic  of  fibrous  materials,  and  the 
resistance  of  a  pillar  to  buckling  is  always  less  than  its  resist- 
ance to  direct  crushing,  and  is  independent  of  length. 


Fig.  336.      Fig.  337.      Fig.  338. 


rlt  tube,  the  H        Thin  malleable  plates  usually  fail  by  the  bending,  pucker- 


-  f. 


i' 


5i6 


THEORY  OF  STRUCTURES. 


'Si  ! 


ing,  wrinkling,  or  crumpling  up  of  the  fibres,  and  the  same 
phenomena  may  be  observed  in  the  case  of  timber  and  of  long 
bars.  >'-a   ■  .-  ■  ■  .-•     "-•   ,■ 

Long  plate  tubes,  when  compressed  longitudinally,  first 
bend  and  eventually  fail  by  the  buckling  of  a  short  length  on 
the  concave  side. 

The   ultimate  resistance  to  buckling   of  a  well-made  and 
well-shaped  tube  is  about  27,000  lbs.  per  square  inch  section  of 
metal,  which   may  be  increased  to  33,000  or  36,000  lbs.  per 
square  inch  by  dividing  the  tube  into  two  or  more  compart 
ments. 

A  rectangular  wrought-iron  or  steel  tube  offers  the  greatest 
resistance  to  buckling  when  the  mass  of  the  material  is  con- 
centrated at  the  angles,  while  the  sides  consist  of  thin  plates 
or  lattice-work  sufficiently  strong  to  prevent  the  'bending  of 
the  angles. 

Timber  offers  about  twice  the  resistance  to  crushing  when 
dry  that  it  does  when  wet,  as  the  presence  of  moisture  dimin- 
ishes the  lateral  adhesion  of  the  fibres. 

5.  Uniform  Stress.— Let  a  short  pillar  be  subjected  to  a 
pressure  of  IV  lbs.  uniformly  distributed  over  its 
end  and  acting  in  the  direction  of  its  axis. 

Let  .5)  be  the  transverse  sectional  area  of  the  pil- 
lar. 


W-' 


W 
Let/>=  -^  be  the  intensity  of  stress  per  unit 

of  area  of  any  transverse  section  AB. 

Let  A'B'  be  any  other  section  of  area  S',  in- 
clined to  the  axis  at  an  angle  6.     The   intensity  of  stress  per 

WW.,, 

/sin  6,  which  may  be 


Fig.  339. 


unit  of  area  oi  A'B'  =  -—  = 


sin  6^ 


S'         S 

resolved  into  a  component/  sin"  6  normal  to  A'B',  and  a  com 

sin  2ff 


ponent/  sin  0  cos  6,  i.e.,/ 


-,  parallel  to  A'£  .     The  last 


intensity  is  evidently  a  maximum  when  6  =  45°,  so  that  the 
plane  along  which  the  resistance  to  shearing  is  least,  and  there- 
fore along  which  the  fracture  of  a  homogeneous  material  would 
tend  to  take  place,  makes  an  angle  of  45°  with  the  axis. 


li^'i; 


UNIFORMLY   VARYING  STRESS, 


517 


None  of  the  materials  of  construction  are  truly  homo- 
geneous, and  in  the  case  of  cast-iron  the  irregularity  of  the 
texture  and  the  hardness  of  the  skin  cause  the  angle  between 
the  plane  of  shear  and  the  direction  of  the  thrust  to  vary 
rom  32°  to  42°.  Brick  chimneys  sometimes  fail  by  the  shear- 
ing of  the  mortar,  the  upper  portion  sliding  over  an  oblique 
plane.    '  ■ 

Hodgkinson's  experiments  upon  blocks  of  different  mate- 
rials led  him  to  infer  that  the  true  crushing  strength  of  a  ma- 
terial is  obtained  when  the  ratio  of  length  to  diameter  is  at 
least  i^;  for  a  less  ratio  the  resistance  to  compression  is  un- 
duly  increased  by  the  friction  at  the  surfaces  between  which 
the  block  is  crushed. 

6.  Uniformly  Varying  Stress. — The  load  upon  a  pillar  is 
rarely,  if  ever,  uniformly  distrib- 
uted, but  it  is  practically  sufficient 
to  assume  that  the  pressure  in 
any  transverse  section  varies  uni- 
jornily. 

Any  variable  external  force  ap- 
plied normally  to  a  plane  surface 
AA  of  area  5  may  be  graphically 
represented  by  a  cylinder  AABB, 
the  end  BB  being  the  locus  of  the 
extremities  of  ordinates  erected 
upon  A  A,  each  ordinate  being  pro- 
portional to  the  intensity  of  press- 
ure at  the  point  on  which  it  is 
erected. 

Let  P  be  the  total  force  upon  A  A,  and  let  the  line  of  its 
resultant  intersect  A  A  in  C;  Cis  the  centre  of  presstire  of  A  A, 
and  the  ordinate  CC  necessarily  passes  through  the  centre  of 
gravity  of  the  cylinder. 

Again,  the  resultant  internal  stress  developed  in  AA  is  P, 
and  may  of  course  be  graphically  represented  by  the  same 
cylinder  yi/4v95.     ,  "     #  '   - 

Assume  that  the  pressure  upon  A  A  varies  uniformly ;  the 
surface  BB  is  then  a  plane  inclined  at  a  certain  angle  to  AA. 


Fig.  340. 


k 


B,U  ■■ 


riir 


*1' 

i 

'i 


!»il 


518 


THEORY  OF  STRUCTURES. 


Take  (?,  the  centre  of  figure  of  AA,  as  the  origin,  and  AA 
as  the  plane  o{  x^y.  '         . 

Let  0  Y,  the  axis  of  jy,  be  parallel  to 
that  line  ££  of  the  plane  BB  which  is 
parallel  to  the  plane  AA. 

Through  EE  draw  a  plane  DD  par- 
allel to  AA,  and  form  the  cylinder 
A  ADD. 

The  two  cylinders  ^^i5^  and  AADD 
are  evidently  equal  in  volume,  and  Of, 
the  average  ordinate,  represents  the  mean 
pressure    over  AA  ;   let    it   be   denoted 

byA- 

At   any   point   R  of  the  plane  AA, 
erect  the  ordinate  RQP,  intersecting  the 
planes   DD,  BB,  in   Q   and   P,    respect- 
ively. 
^"''  341-  Let  x,y  be  the  co-ordinates  of  R. 

The  pressure  at  R 


=  p  =  PR  =  PQ  +  QR  =  PQ-\.OF=ax+p^ 


0  > 


a  being  a  constant  depending  upon  the  variation. 

Note. — The  sign  of  x  is  negative  for  points  on  the  left  of  0, 
and  the  pressure  at  a  point  corresponding  to  R  xs  p^  —  nx. 

Let  x^,  y^  be  the  co-ordinates  of  the  centre  of  pressure  6. 

Let  AS  be  an  elementary  area  at  any  point  R. 

Then  pAS  is  the  pressure  upon  AS,  and  2{pJS}  is  the 
total  pressure  upon  the  surface  AA,  2  being  the  symbol  of 
summation. 

Hence, 


x,2{pAS)  =  2(pxAS),    and    y,S(pAS)=^ 
■    Butp=p^-^ax. 
;  •  •  .-.  x,2{{p,  +  ax)AS\  =  2{(p,x  +  ax*)AS} 


o). 


and 


.y.^KA  +  ax)AS\  =  2\{p,j>  +  axy)AS\. 


UNIFORMLY   VARYIiVG  STRESS. 


5'9 


Now  0  is  the  centre  of  figure  of  A  A,  and  therefore  "S^ixAS) 
and  'H^^yAS)  are  each  zero. 

Also,  ^X'^'^)  =  -^f  ^X-*'"^-^)  is  the  moment  of  inertia  (/)  of 
A  A  with  respect  to  OY,  and  "^{xyAS)  is  the  product  0/  inertia 
[K)  about  the  axis  OZ. 


.'.  x„p,S=  aI=x^P 


0) 


and 


yJ,S=^aK  =  y,P. (2) 


Cor.  I.  In  any  symmetrical 
section  y^  is  zero,  and  x,  is  the 
deviation  of  the  centre  of  pres- 
sure C  from  the  centre  of  fig- 
ure 0. 

Let  x^  be  the  distance  from 
0  of  the  extreme  points  A  of 
the  section. 


|A  A 


—9-X  » 

0  "C 


Fig.  34a. 


The  greatest  stress  in  A  A  is/^  -+-  ax^  =/, ,  suppose. 
Buta=^^,by  eq.  (1). 

•  •AH 7 —  —Pi* 


or 


A 

P. 


i  +  ^S 


(3) 


It  is  generally  advisable,  especially  in  masonry  structures, 
to  limit  x„  by  the  condition  that  the  stress  shall  be  nowhere 
negati'i'e,  i.e.,  a  tension.  Now  the  minimum  stress  is/,  —  ax^ , 
so  that  to  fr  '1  this  condition, 

/>f  >  or  =  ax, .     But/,  =  ax,  -\-p, ;     .-. /,  <  or  =  2/». 


ii 


■'^^tiiujA^i^fc' 


a 


i! 


<ii 


I   i 


520  THEORY  OF  STRUCTURES. 

Hence,  by  eq.  (3), 


— -  <  or  = » 

2/„  .    XA\ 


I  +-7-5 


and  therefore 

x^x,S 


J      <  or  =  I  ;      I.e.,  x^  <  or  =  ^, 


Cor.  2.    The  uniformly  varying  stress  is  equivalent  tu  a 
single  force  /^  along  the  axis,  and  a  couple  of  moment 


PXC0  =  P  Vx,"  +  f-'  r=  a  Vf  +  K\ 
Cor.  3.  The  line  CO  is  said  to  be  conjugate  to  OY. 


If  the  angle  COX  =  6,     then 


,  (4        ^^        ^ 
cot  (/  =  —  =  --. 


\  7.  Hodgkinson's  Formulae  for  the  Ultimate  Strength  of 
Long  and  Medium  Pillars. — WHien  Along \^\\\ax  is  suljjccii.'d  tu 
a  crushing  force  it  first  yields  sideways,  and  eventually  breaks 
in  a  manner  apparf '.ly  similar  to  the  fracture  of  a  beam  unclcr 
a  transverse  load.  Thi.s  similarity,  however,  is  modified  by  the 
fact  that  an  initial  longitudinal  compression  is  induced  in  tlic 
pillar  by  the  superimposetl  load. 

Ilodgkmson  deduced,  experimentally,  that  the  streiii,^tli  of 
long  solid  round  iron  and  square  timber  pillars,  ^vitli  fiat  and 
firmly  bedded  ends,  is  given  by  an  expression  of  the  form 


W=^A 


d" 


tm^ 


W^  being  the  breaking  weight  in  tons  of  2240  lbs. ; 
d        "         "    diameter  or  side  of  the  pillar  in /;/f/i!^i  / 
/         "        "    length  of  the  pillar  in  /dV/; 
«  and  ;;/  being  numerical  indices  ; 


FCKMUI..K   FOR    ULTIMATE   STNEiXCTH  OF  PILLARS.     $21 

A   bcin-;   a   conslaiit  varyiii^f  uitli  the  material   and  with   the 
sectional  form  of  the  pillar. 

For  iron  pillars «  =  3.6  and  in  =.  1.7 

"  timber  pillars «  —  4     and  »i  =  2 

"  cast-iron A  —  44. 16 

"  wrouijht-iron A  =  \  33.75 

"  dry  Dantzic  oak A  =  10.95 

"  dry  red  deal A  =  7.8 1 

"  dry  French  oak A  =  6.9 

The  strength  of  /o//^'-  holloxv  round   cast-iron    pillars  was 
found  to  be  given  by 

W=  44.34 jY^. , 


(/  being  the  external  and  d,  the  internal  diameter,  both  in 
inches. 

Thus,  the  strength  of  a  hollow  c  ist-iron  pil'  ir  is  approxi- 
nial'jly  equal  to  the  difference  between  the  strengths  of  two 
solid  cast-iron  pillars  whose  diameters  are  equal  to  the  external 
and  internd  diameters  of  the  hollow  pillar. 

The  strength  of  uiediiun  pillais  may  be  obtained  by  the 
formula 


W  = 


IVfS 


IV  being  the  breaking  weight  in  tons  of  2240  lbs. ; 

II'       "      "  "  '      '•      "       "     as    derived 

from  the  formula  foi-  /o>ij^  pillars  ; 
/being  the  ultimate  crushing  strength  in  tons  per  square  inch  ; 
6"  being  the  sectional  area  of  the  pillar  in  square  inches. 

Again,  if  the  ends  of  a  cast-iron  pillar  are  rounded,  the 
above  formulae  may  be  still  employed  to  determine  its  strength, 
A  being  14.9  for  a  solid  and  13  for  a  hollow  pillar. 


( i 


522 


THEORY  OF  STRUCTURES. 


'iK 


\\\ 


8.   Gordon's  Formula  for  the  Ultimate  Strength  of  a 

Pillar. — The  method  discussed  in  the  preceding  articles,  being 

practically  very  inconvenient,   is    not   generally  used, 

and  the  present  article  will  treat  of  Professor  Gordon's 

formula,  which  has  a  better  theoretical   basis  and  is 

easier  of  application. 

The  effect  of  a  weight  f^upon  a  pillar  of  lengtli  / 

and  sectional  area  S  may  be  divided  into  two  parts : 

{a)  A  direct  thrust,  which  produces  a  uniform  com- 

.       W 
pression  of  intensity  -rr  .—  p^ . 

(^)  A  bending  moment,  which  causes  the  pillar  to 
yield  in  the  direction  of  its  least  dimension  (/;). 

Let  J/  be  the  greatest  deviation  of  the  pillar  from 
the  vertical. 

The  bending  moment  M  at  the  point  of  maximum  stress 
may  be  represented  by  Wy. 

Let  /,  be  the  stress  in  the  extreme  layers  due  to  this  bend- 
ing moment. 


Now 


M=^-^I  =  txpJ>h\ 


c  being  the  distance  of  the  layer  under  consideration  from 
the  neutral  axis,  /<  a  constant  depending  upon  the  sectional 
form,  and  b  the  dimension  perpendicular  to  the  plane  of  flexure. 


.*.  yw////'  =  Wy,    and    p^<x 


Wy_ 
bh'' 


r 


But^ocr-.   (Art.  9,  Chap.  VI.)] 


,    wr    ivr    ^r      ,  /• 


VALUES  OF  a  AND  /    GORDON'S  FORMULA. 

a  being  some  constant  to  be  determined  by  experiment. 
Hence,  the  total  stress  in  the  most  strained  fibre  is 


523 


/  =  A+A=a(i+«^,), 


or 


W 


f 


/e 


which  is  Gordon's  formula. 

Cor. — If  the  weight  upon  the  pillar  causes  the  stress  in  any 
transverse  section  to  vary  nnifornily,  the  direct  thrust  in  the 

h 


W 


w 

stead  of  -jT,  (Cor.  i,  Art. 


extreme  layers  is  -^ 

6,)  .i\  being  the  greatest  deviation  of   the  line   of  resultant 
thrust  from  the  axis  of  the  pillar. 

Let  k  be  the  radius  of  gyration  of  the  cross-section.     Then 

Sk'  =  /, 

and  the  expression  for  the  direct  thri      may  be  written 


Hence,  Gordon's  formula  becomes 


W 


=  A  = 


/ 


//'  ^  2/6' 


V 

/ 


9.  Values  of  a  and  /. — The  following  table,  giving  the 
values  of  the  constants  a  and/  in  Gordon's  formula,  has  been 


m 


524 


THEORY  OF  STRUCTURES. 


prepared  by  taking  an  average  of  the  best  known  results,  and 
is  applicable  to  round  and  square  pillars  ivith  square  ends. 


For  cast-iron  solid  rectangular  pillars :    80,000 


"  "  "     round  "     

"  "         hollow  rectangular    "     ...    . 

"  "  "       round  "     

For  wrotight-iron  solid  rectangular  pillars . 
"     round  "      , 

"  "  thick  hollow  round  "  . 
For  mild-steel  solid  rectangular  pillars 

"  "  "     round  "     

"  "  hollow  round  "     

For  strong-steel  solid  rectangular  pillars.. . 
"     round  "     ... 

"  "  hollow  round  "     ... 

Tor  fine-timber  solid  rectangular  pillars... . 

"  "  "  rouiiJ  '*  .... 
For  dry  oak  timber 


/in  lbs. 

per  sq.  in. 

80,000 

TSC 

80,000 

ih!\ 

80,000 

Tiini 

So.ooo 

soil 

36,000 

TfffW 

36,000 

^SST! 

36,000 

Il'niT 

67,200 

iTiVij 

67,200 

TTfiiS 

67,200 

•ss'oci 

114,000 

TTlTD 

114,000 

5ui7 

114,000 

TXffT 

5,000 

-u 

5,000 

-u 

7,200 

1 

If  Gordon's  formula  is  applied  to  pillars  with  pin  ends,  4;? 
takes  the  place  of  a  ;  and  if  to  pillars  with  one  pin  end  and  one 
square  end,  ^a  takes  the  place  of  a. 

10.  Graphical  Comparison  of  the  Crushing  Unit  Strength 
of  Solid  Round  Cast-iron,  Wrought-iron,  and  Mild-steel 
Pillars. 

/ 

The  crushing  unit  stress  is  given  hy  p  = j^. 

/ 

Take  the  different  values  of  -j  as  absciss.ne,  and  the  corre- 

n 

spending  values  of  /  as  ordinates ;   the  resulting  curves  are 

shown  in  Fig.  344. 

Hence,  the  strength  of  a  mild-steel  pillar  always  exceeds 

that  of  a  wrought-iron  pillar  but  is  less  than  that  of  a  cast-iron 

/ 
pillar  when  -j  <  10.7  ;  a  wrought-iron  pillar  is  stronger  or  weak 

than  a  cast-iron  pillar  according  as  r  >  or  <  28.5. 


er 


APPLICATIONS  OF  GORDON'S  FORMULA. 


525 


ilts,  and 
(is. 


ISI5 


"s'li) 

soil 

) 

Ti^M 

3          ^5STi 

3         SB^Oii 

0          JTi'il  0 

0 

TTOS 

0 

•ss'di! 

0 

TTrtii 

0 

Huff 

K)         Ts\is 

XJ 

irb 

DO 

uJfi 

DO 

ISU 

1  ends,  4(? 

d  and  one 

Strength 

/I 

ild-steel 

the  corre- 
lurves  are 

exceeds 
cast-iron 

lor  wcakei 


80,000  lbs, 

67,300  lbs, 
ClOfO  lbs, 


II.  Application  of  Gordon's  Formula  to  Pillars  of  other 
Sectional  Forms. 

In  any  section  whatever,  the  least  transverse  dimension  for 
Ccilculation  (i.e.,  h)  is  to  be  measured  in  the  plane  of  greatest 

ricxnre. 

Thus,  it  may  be  taken  as  the  least  diameter  of  the  rectant;le 
circumscribing  ice  (Fig.  345),  chaiuiel  (Y\^.  346\  and  cruciform 
I  Fig.  347)  sections,  and  as  the  perpendicular  from  the  angle  to  the 
opposite  side  of  a  triangle  circumscribing  rtz/^V^^l-'ig.  348)  sections. 

7i, 


i 

f-«-| 

•^ 

K 

iJ 

'S 

l.<  --, 

Fig. 

345- 

Fig.  346. 

Fir, 

3t7- 

Fig.  348. 

From  a  series  of  experiments  upon  wrought-iron  pillars  of 

I 


these  sections, /"was  found  to  be  42,e;oo  lbs.,  and  n, 


900 


In   cast-iron  struts  of  a  cruciform   section  /  —  8o,oCK)  lbs. 


and  a  — 


inii 


400 


:ii 


^26 


THEORY  OF  STRUCTURES. 


Ni  i 


These  results  are  only  approximately  true,  and  apply  to 
pillars  fixed  at  both  ends. 

12.  Rankine's  Modification  of  Gordon's  Formula.— The 

factor  a  in  Gordon's  formula  is  by  no  means  constant,  and  not 
only  varies  with  the  nature  of  the  material,  with  the  length  of 
the  pillar,  with  the  condition  of  its  ends,  etc.,  but  also  with  tiic 
sectional  form  of  the  pillar.  The  variation  due  to  this  latter 
cause  may  be  eliminated,  and  the  formula  rendered  somewhat 
more  exact,  by  introducing  the  least  radius  of  gyration  instead 
of  the  least  transverse  dimension. 
If  k  is  the  least  radius  of  gyration, 

• 

mass       nbh        n     ' 

m  and  n  being  constants  which  depend  upon  the  sectional 
fo»-m,  Thus,  Gordon's  formula  for  pillars  with  square  ends 
may  be  written 


W 


=  A  = 


/ 


/»» 


in  which  a^  is  independent  of  the  sectional  form,  all  variations 
of  the  latter  being  included  in  k^.  This  modified  form  of 
Gordon's  formula  was  first  suggested  by  Rankine. 

4^',  is  substituted  for  «,  if  the  pillar  has  two  pin  ends,  and 

-a^  or  2rt,  is  substituted  for  a,  if  the  pillar  has  one  pin  end  and 

one  square  end. 
Rankine  gives 


for  wrought  iron,      /=  36000  lbs., 
for  cast-iron,  '/=  80000  lbs., 

for  dry  timber,  /=    7200  lbs.. 


I 

I 


=  36000; 
=  6400; 
=  3000; 


FAN  KIKE'S  MODIFICATION  OF  GORDON'S  FORMULA.     527 

In  good  American  practice  the  safe  working  unit  stress  in 
bridge  compression  members  is  determined  by  the  formula 


Safe  vvorkiner  unit  stress  = 


/' 


f  being  8000  lbs.  for  wrought-iron  and   10,000  lbs.  for  steel, 
and  -  being  40,000  for  two  square  ends,  30,000  for  one  square 

and  one  pin  end,  and  20,000  for  two  pin  ends. 
Another  formula  often  employed  is. 

Working  stress  in  lbs.  per  sq.  in.  X  (4  +  ^)  =  i  -\-xH' ' 

If  bt-'ing  the  ratio  of  length  to  least  breadth,  where,  in  the  case 
of  wrought-iron, 


/' 
/' 
/' 


38,500  lbs.  and  - 


38,500   "     " 

37,800  "    " 


5820  for  two  square  ends; 


3000    "   one  square  and  one  pin  end. 


—  igoo    "    two  pin  ends. 


Tht  factor  of  safety,  viz.,  4  -j ,  increases  with  H,  and  par- 
tially provides  for  the  corresponding  decrease  in  the  strength 
to  resist  side  blows. 

Examples. — According  to  Rankine  the  ultimate  compres- 
sive strength  of  wrought-iron  struts,  in  pounds  per  square 
inch,  is 


36000 


■    1  + 


I     r 
36000^ 


m 
I  ill 


:.^i^.Jtf!iU-v  luil 


• 


: 


m 


w 


111        it 


i  i  V 


528 


THEORY  OF  STRUCTURES. 


h' 


If  the  section  is  a  solid  rectangle,  1^  —  — ,  and  hence 

36000 


A  = 


'    3000  /[' 


le 


If  the  section  is  a  solid  circle,  k^  =  -z,  and  hence 

36000 


A  = 


1  + 


I     /'• 


2250 /t" 


le 


If  the  section  is  a  thin  annulus,  k^  =  -^,  nearly,  and  hence 

36000 


A  = 


1  + 


I     /' 


4500  A' 


I . 


Cor, — If  T  is  small,   W  ■=■  fS. 

1  f^h^  fT 

If  ^  is  large,   {^= —  .-=_. 

Comparing  the  last  result  with  eq.  (5),  Case  4,  Art.  16, 

I       ^Eit' 


tty 


f  ' 


which  gives  a  theoretical  value  of  a, ,  the  actual  value  being 

somewhat  different. 

13.  Values  of  h^  for  Different  Sections. 

/     A' 
{a)  Solid  rectangle :  1^  ■=—■=.  —,  h  being  the  least  dimen- 

O         12 

sion. 

{b)  Hollow  rectangle:  k*  =  -^  =  —  f  , .  _  ,/./),  b,  h  being 

the  greatest  and  least  outside  dimensions,  and  b\  h'  the  great* 
est  and  least  inside  dimensions,  respectively. 


m 


;■' ! 


VALUES  OF  k*  FOR  DIFFERENT  SECTIONS. 

Let  /  be  the  thickness  of  the  metal.     Then 
b'  t=b  —  2t    and     li  =  h  —  2t, 


529 


and  hence 


I  bh'  -{b-  2t){h  -  2ff       h'  zb-\-h 


^'  ~  12  bh-{b-  2t){h  -  2t)   ~  12  b-\-/? 

approximately,  when  /  is  small  compared  with  /i,  i.e.,  for  a  i/u'n 
hollotv  rectangle. 

For  a  square  cell,  k'  =  -^. 

(c)  Solid  triangle :  k^  =  -^,  =  — ^,  k  being  the  height. 

(d)  Hollow  triangle  :  k"  =  ^^  =  -^  , .  _  ,,.,  ,b,  h  being  the 

base  and  height  of  the  outside  triangle,  and  b\  Ji!  the  base  and 

b       h 
height  of  the  inside  triangle,  respectively.     Also,  -r,  =  -,-. 


: 
[ 

•i 

'     I 


Hence,  for  a  t/tin  triangular  cell,  k'  =  — . 

/       li' 
(e)  Solid  cylinder :  k'  =  -  =  —^,  h  being  the  diameter. 

/       n 
(/)  Hollow  cylinder :  k'  =  T.  =  ~fi(^''  "^  ^''' ")'  ^^  '^"^  ^^'  '^^'"g 

the  external  and  internal  diameters,  respectively. 

Hence,  for  a  t/nn  cylindrical  cell,  k''  =  -—,  approximately. 

o 

Example. — Gordon's  formula  for  hollow  cylindrical  cast- 
iron  pillars  is 


W 


=  A  = 


/ 


/ 


^    '    500  /:'       ^    '    4000  k* 


'i.aci^f  I     .^ 


UtM^^ 


MiiiM 


lU 


!*M  !! 


^530  •  "theory  OF  STRUCTU^AS. 

The  relation  /,  = — 77  may  be  assumed  to  hold  for 

'   4000  k^ 
hollow  square  struts  and  also  for  struts  of  a  cruciform  section. 

Ex.  I.  For  a  hollow  square  having  its  diagonal  equal  to 
the  internal  diameter  of  the  hollow  cylinder,  i.e.,  h', 


f 


6   =77'  ^"'^  f^=-~~iri" 


1  + 


1000 //'* 


1  -fl 


ir 


Ex.  2.  If  the  side  of  the  square  is  equal  to  the  external 
diameter,  i.e.,  h,  then 


'^^  ~  "6  '     ^"^    ^' 


/ 


1  + 


3     /'• 


2000  h^ 
(g)  Cruciform  section,    the    arms  being  equal : 


V 

b 

^1 

I 

Fig.  349. 


,         M'    ,    hb'  b'  ^  ,.  ,, 

/_ L — ;    5  =  2bh  —  fV. 

12  '12       12 


I  bh'  +  hb'-b'       h' 

.*.  <fe  = h n —  =  — »  nearly. 

12      2bh  —  h^  24  -^ 


Hence,  the  formula  for  a  cast-iron  pillar  of  cruciform  section 
may  be  written 


W 


/ 


/ 


1  + 


I      /' 


3    I' 


4000  k^ 


500  A* 


{li)  Angle-iron  of  unequal  ribs,  the  greater  being  b  and  the 
less  h'. 

^'  =  7^<^'  -[-  ir  approximately, 


mmmr 


VALUES  OF  k*   FOR  DIFFERENT  SECTIONS. 


531 


Hence,  if  ^  =  ^,  i.e.,  if  the  ribs  are  equal,  k^  = 


24' 


(t)  Channel-iron,  the  dimensions  being  as  in  Fig.  350 


^  _  bt'  +  2h't      2bht\h  +  /)' 


12 

2ht 


42ht  +  bt) 
2b/it^ 


,,  (  2^/  ,        2bhf       ) 


Also, 


4{2ht  +  bt) 
S=:  bt-\-  2ht. 

2ht 


1 

t            1 

* 

— ft— 

A 

• 

Fig.  350. 


.-.  k'  =  /i'  I  Y 


2Mi" 


2(2/4/  +  bt)  '  4(2/^/  +  bt) 


4- 


Let  the  area  of  the  two  flanges  z=.  A  ■=  2ht,  and  let  the 
area  of  the  web  ^  B  =  bt.     Then 


k^  =  k^[^ 


+ 


AB 


2{A  +  By  4{A-^B) 


>')'  I  • 


{k)  H-iron,  breadth  of  flanges  being  b,  length  of  web  k,  and 
thickness  of  metal  / : 

,        b't  ,  he         b't  ,  o         z     ,    r 

7=2 — H =  2  — ,  nearly  ;      5=2W  +  ^/. 

12  '    12         12  "^  ' 


•.  /fe'  =  - 


2(5/ 


\2  2bt-\-ht      12A  +  B' 

A  being  the  area  of  the  flanges,  and  B  the  area  of  the  web. 
(/)  Circular  segment,  of  radius  r  and  length  rO  : 


k*  =  r* 


1  sm  t'      ^        2 

2  +  "2^  6*^^ 


Hence,  for  a  semicircle,  since  6  =z  tt, 

,&»  =  r«  {  -  -  ^  I  =  ^  ,  nearly. 


^jSj 

li 

'■  <■ 


If 


I' 


V 


M 


;|::!l 


'Hi 

liti 

•i  ■,! ': f 

if 


i 


I    II 


532 


THEORY  OF  STRUCTURES. 


(;«)  Barlcnv  rail:   k'  =■  — ,  nearly. 

(»)  Two  Barlow  rails,  riveted  base  to  base :  k*  =  'ijlr\ 
nearly. 

14.  American  Iron  Columns. — In  1880  Mr.  G.  Bouscaren 
read  before  the  American  Society  of  Civil  Engineers  a  paper 
containing  the  results  of  a  series  of  experiments  made  for  the 
Cincinnati  Southern  Railroad  upon  Keystone,  square,  Phoenix, 
and  American  Bridge  Co.'s  columns. 


KEYSTONE 
Fig.  351. 


AM.  BRIDQE  CO, 


SQUARE  PHOENIX 

F>G-  352.  Fio.  353.  Fig.  354. 

These  experiments  show,  as  those  of  Hodgkinson  and 
others  have  also  shown,  that  the  strength  of  iron  and  steel 
columns  is  not  only  dependent  on  the  ratio  of  length  to  diam- 
eter, and  on  the  form  of  the  cross-section,  but  also  on  the 
proportions  of  parts,  details  of  design  and  workmanship,  and 
on  the  quality  of  the  material  of  which  the  columns  are  con- 
structed. 

Further,  they  seem  to  lead  to  the  conclusions  that  Gordon's 
formula  is  more  correct  as  modified  by  Rankine,  and  that,  in 
the  case  of  columns  hinged  at  both  ends,  Rankine's  formula, 
with  rt,  assumed  at  double  the  value  it  has  when  the  formula  is 
applied  to  columns  with  flat  ends,  is  practically  correct. 

The  subjoined  table  gives  the  values  of  the  constants 
a^  and  f  as  deduced  from  Bouscaren's  experiments  by  Prof. 
W.  H.  Burr. 

In  1 88 1  Messrs.  Clarke,  Reeves  &  Co.  presented  to  the 
American  Society  of  Civil  Engineers  a  paper  containing  the 
results  of  experiments  upon  twenty  Phoenix  columns,  which 
appeared  to  show  that  neither  Gordon's  nor  Rankine's  formula 
expressed  the  true  strength  of  a  column  of  the  Phoenix  type. 
In  the  di.scussion  that  followed  the  reading  of  this  paper,  how- 
ever, it  was  demonstrated  that,  within  the  range  of  the  experi- 
ments, the  strength  of  intermediate  lengths  and  sections  of 


AMERICAN  IRON  COLUMNS. 


533 


=  .393'-'. 

ouscaren 
a  paper 

e  for  the 
Phoenix, 


gi."  •.'^ 


kM.  BRIDQECO. 
FiC.  354- 

inson  and 
and  steel 
h  to  diam- 
Iso  on  the 
.nship,  and 
IS  are  con- 
It  Gordon's 
ind  that,  in 
s  formula, 
formula  is 

;ect. 
constants 

s  by  Prof. 

ted  to  the 
gaining  the 
ins,  which 
■e's  formula 
Icenix  type. 
^aper,  how- 
[the  experi- 
Isections  of 


For  keystone  columns  with  flat  ends — swelled 

"  "  "       "       "    —straight      (open      or 

closed) 

•'  "  "  "       "       "  —open  (swelled  straight) 

"  "  "  "       pin  ends — swelled 

For  square  columns  with  flat  ends 

"  "    pinends 

For  Phoenix  columns  with  flat  ends 

"  "  "  "     round  ends 

"     pinends 

For  American  Bridge  Co.'s  columns   with  flat  ends 

"          "            "     round  ends.... 
"  "  "  "     pinends 


/  in  lb>. 


36,000 

39.500 
38,300 
38.300 
39,000 
39,000 
42,000 
42,000 
42,000 
36,000 
36,000 
36,000 


"I 


msoo 

issoo 

1 
mooo 

1 

ttooo 

1 

17000 

\ 

(0000 

1 

11600 

1 

SITOO 

1 

4«000 

1 

11*00 

1 
titoo 


Phoenix  columns  can  be  obtained  either  from  Rankine's  for- 
mula by  slightly  changing  the  constants,  or  from  very  simple 
new  formulae. 

I 


Mr.  W.  G.  Bouscaren  showed  that  by  making  «,  = 

W 


lOOOOO 


and/=  38000,  the  calculated  values  of  -^  agree  very  nearly 

with  the  actual  experimental  results. 

Mr.  D.  J.  Whittemore  gave  the  following  (only  applicable 
for  lengths  varying  from  5  to  45  diameters)  as  expressing  the 
probable  ultimate  strength  of  these  columns  : 


W^  lbs.  =  (1200  —  H)io-\- 


525000 


//being  the  ratio  of  length  to  diameter. 

Mr.  C.  E.  Emery  stated  that  the  ultimate  strength  in  each 
case  is  approximately  represented  by  the  formula 


^Tlbs.  = 


355063  +  30950// 


i^+6.175 
//being  the  ratio  of  length  to  diameter. 


i 


r 


ii 

;  1 

In 

1 

II 

aia.:>^'';';:  S^'ji^i 


S84t 


THEORY  OF  STRUCTURES. 


Taking  the  different  values  of  //  as  abscissai,  and  of  W  as 
ordinates,  this  is  the  equation  of  an  hyperbola.  It  agrees  very 
accurately  with  the  experimental  results  from  20  diameters 
upwards;  at  15  diameters  the  calculated  values  of  W  are 
greater  than  those  given  by  the  exf  oriments  ;  for  a  less  num 
ber  of  diameters  the  experimental  results  are  the  higher,  but 
the  variations  are  slight,  and  are  provided  for  in  the  factor  of 
safety. 

The  following  very  simple  formuL-e,  due  to  Prof.  W.  H. 
Burr,  give  results  agreeing  closely  with  those  obtained  in  the 
experiments : 

/ 
For  values  of  -r   ^  30,  the  ultimate  strength  in  pounds  per 

square  inch 


=  64700  —  4600, 


\fi- 


For  values  of  -r  between  30  and  140,  the  ultimate  strength 
in  pounds  per  square  inch 

=  39640-46^, 


k  being  the  radius  of  gyration. 

15.  Long  Thin  Pillar. — Let  ACB  be  the  bent  axis  of  a 
,B        thin  pillar  of  length  /,  having  two  pin  ends  and  carry- 
ing a  load  W  at  B. 

Let  d  be  the  greatest  deviation  of  the  axis  from  the 
vertical.     Then 

E 
Wd  —  bending  moment  =  -h/,     .    •    (0 

—  being  the  curvature  of  the  pillar  and  /the  moment 

of  inertia  of  the  most  strained  transverse  section. 
This  equation  is  only  true  on  the  assumptions  that— 
(i)  initially,  the  pillar  is  perfectly  straight; 
(2)  initially,  the  line  of  action  of  the  load  coincides  with  the 
axis  of  the  pillar ; 


wm 


LONG   THIN  PILLAR. 


535 


(3)  the  material  of  the  pillar  is  homogeneous. 

These  assumptions  cannot  be  fulfilled  in  practice,  and  varia- 
tions from  theoretical  accuracy  may,  perhaps,  be  provided  for 
by  supposing  that  the  line  of  action  of  the  load  is  at  a  small 
distance  x  from  the  axis  of  the  pillar.  The  bending-moment 
equation  then  becomes 


^n^+^)  =  |/=7A 


(2) 


/,  being  the  skin  stress  due  to  bending  at  a  distance  c  from  the 
neutral  axis. 

Again,  assuming  that  the  bent  axis  is  in  the  form  of  an  arc 
of  a  circle, 

I      U  .  . 

R^-P (3) 

.'.W{d^x)=.%EIj^^^p,     ....    (4) 

and  consequently 

Wx 
^^-p~Zw ^5) 

where 

p='^ (6) 

If  the  line  of  action  of  the  load  W  coincided  with  the  axis 
of  the  pillar,  then  x  would  be  nil. 

Hence,  by  eq.  (5),  so  long  as  the  load  is  less  than  /',  d  =  o, 
and  the  failure  of  the  pillar  would  be  due  to  direct  crush- 
ing.   If  the  load  is  equal  to  P,  d?  would  become  indi  terminate 

(=-)  and  the  pillar  would  remain  in  a  state  of  neutral  equi- 
librium at  any  inclination  to  the  vertical. 

It  is  impossible  that  W  should  exceed  P,  as  d  would  then 
be  negative  ;  and  therefore  a  load  greater  than  P  would  cause 
the  pillar  to  bend  over  laterally  until  it  broke. 


-'W   ij»^ 


VA] 


Mi 


f 

1' 

1 

1 

f  n 

r' 

^ 

ML.      V 


536 


THEORY  OF  STRUCTURES. 


Thus,  P  —  -yr  must  be  the  theoretical  maximum  compres- 
sive strength  of  the  pilar. 

Again,  let  A  be  the  area  of  the  section  under  consideration  ; 
"    p  be  the  total  intensity  of  the  skin  stress  at  the 
section  ; 

"   f  be  the  intensity  of  the  direct  stress  due  to  W 

_W_ 

~  A  ' 
"    /",  be    the    intensity   of    the    stress    due    to  P 

__  P_ 

~A' 


Then 


P=f±A  =  ^±W{d+x)j (7) 


the  sign  of  /,  being  positive  for  the  coinpressed  side  of  the 
pillar  and  negative  for  the  side  in  tension. 

.•./  =  -^'(i±(«r+^)4^)=/(i±(^+^)|5),  '.  (8) 

k  being  the  radius  of  gyration. 

Let  h  be  the  least  transverse  dimension  of  the  section  in 
the  plane  of  flexure.     Then 

c  <x  h    and     k  also  oc  lu 


c       n 


«  being  a  coeflficient  depending  upon  the /^r;«  of  the  section. 
For  a  rectangle,  «  =  6  ;  for  a  circle,  «  =  8 ;  also, 

Px 


section  in 


LONG    THiN  PILLAR. 


537 


Thus,  however  small  x  may  be,  p  continually  increases  as 
the  difference  between  /,  and  /  diminishes.  The  pillar  will 
therefore  fail  for  some  value  of  p  less  than  the  theoretical 
maximum.  This  is  in  accordance  with  experience,  as  it  is 
found  that  a  small  load  causes  a  moderate  flexure  in  a  long 
pillar,  and  that  this  flexure  gradually  increases  with  the  load 
until  fracture  takes  place. 

In  no  case  should  /  exceed  the  clastic  limit,  as  in  such 
case  a  set  would  be  produced  and  the  deviation  x  would  be 
increased. 

If  the  tensile  strength  of  the  material  of  the  pillar  is  small, 
as  in  the  case  of  cast-iron,  failure  may  arise  from  the  tearing  of 
the  stretched  layers. 

Cor.  I.  The  above  also  applies  to  the  case  of  a  pillar  with 
one  end  fixed  and  the  other  free,  but  the  value  of  P  is 
EI 

Cor.  2.  According  to  Euler  (see  following  article),  the 
more  correct  value  of  P  is  /xEIji,  /^  being  I,  2,  J,  or  4,  accord- 
ing as  the  pillar  has  two  pin  ends,  one  fixed  end  and  one  end 
guided  in  the  direction  of  the  thrust,  one  fixed  and  one  free 
end,  or  two  fixed  ends. 


P  evidently  <x  -j^  ex  EA-p  oc  EA 


§■ 


Hence,  (a)  the  strength  of  a  long  pillar  is  proportional  to 
the  coefficient  of  elasticity ;  {d)  the  strengths  of  similar  pillars 
are  as  the  sectional  areas. 

Again,  /,  <x  -^  oc  d. 

But  Wd  =  =^'-/ oc /,  oc  d. 

Hence  W  is  approximately  constant,  and  the  weight  which 
produces  moderate  flexure  is  approximately  equal  to  the  break- 
ing weight. 

Example. — Find  the  crushing  load  of  a  solid  mild-steel 
pillar  3  in.  in  diameter  and  10  ft.  long,  with  two  pin  ends. 


I'M 
it '  '  ' 

-  ■      i 

1 

■: 

i  > .  i  i 

|! 

'     i 

|; 

:1     ' 

1 

i. 

|lK 

H 

;■ 

H 

i 

|»W 

i 

IH 

■] 

1 

1 

m 

; 

H 

i 

538 


THEORY  OF  STRUCTURES. 


Also  find  the  deviation  (-r)  of  the  line  of  action  of  a  load  of 
20,000  lbs.  from  the  axis  of  the  pillar,  sc  \hat  the  maximum 
intensity  of  stress  may  not  exceed  10,000  lbs.  per  square  inch. 

By  Gordon's  formula  and  the  table,  page  524, 


the  crushing  load 


_      672oo;r .  | 

I  _i_      4    /izov 


85292.3  lbs. 


Again,  the  theoretical  maximum  compressive  strength  P 
SEI      8  X  28000000  7r(3)* 


/' 


(120)' 


64 


=  61875  lbs. 


/. 


P-  w-f,-f 


6i_875 
41875 


99 
67- 


Hence 


20000/       ,    QQ 
I0000  =  -^^^I  + 


'^-\ 


67 


8   ^ 

X 

3 


)' 


ox  X  =  .65  in. 
16. 


9 


M 


L 


Long  Columns  of  Uniform  Section.   (  Pluler's  Theory.) 

Case  I.  Columns  wiih  both  ends  hinged. 

— The   column   OA    of  length   /  is   bent 

under  a  thrust  P  and  takes   the   curved 

form  OMA. 

Take  O  as  the  origin,  the  vertical 
through  O  as  the  axis  ^^i  x.  and  the  hoii 
zontal  through  O  as  the  axis  oi  y. 

Consider   a  section   at   any   point  M 

{x,y).     If  there  is  equilibrium  and  if  the 

line   of  action   of   P  coincides  with  the 

-""-"-^yj      '     axis  of  the  column,  the  equation  of  mo- 

FiG.  356.  ments  at  M  is 


or 


1? 


EI 


y  = 


a*y. 


^0 


LONG   COLUMNS  OF   UNIFORM  SECTION. 


539 


dy 


Multiplying  each  side  of  the  equation  by  -j-  and  integrating, 

(!)■  =  «•{*•-/), (2) 


h  being  a  constant  of  integration 

dy 


•  \/b^ 


=  adx. 


Integrating, 


sin"'  ( -^j  =  ax  -|-  Cy 


or 


y  ^=  b  s\n  {ax  -f-  c), 


(3> 


c  being  a  constant  of  integration. 

When  X  =  o,  y  \s  also  o,  and  hence  ^  =  o  or  t  =  o. 

\[  b  =^  o,  y  \s  always  o,  and  lateral  flexure  is  impossible., 

Take  c  =  o.     Then 


y  =  b  s\n  ax.      .... 
Also,  when  x  —  OA  =  OMA ,  nearly,  =  l,y  z=o. 

.'.  O  =  b  sin  al, 


{^1 


ot 


nn  =  al 


=V^' 


and  hence 


TT" 


P^n'EI-ji, 


(5) 


Now  the  least  value  of  P  evidently  corresponds  to  «  =  i, 
and  hence  the  minimum  thrust  which  will  bend  the  column 
laterally  is 

P^EI~. 


wm- 


540 


THEORY  OF  STRUCTURES, 


Cor.  I.  If  the  column  is  made  to  pass  through  N  point: 
dividing  the  vertical  OA  into  N  A,-  \  equal  divisions,  then 

J/  =  o  when  x  = 


and  therefore,  by  eq.  (4), 

o  =  ^  sin 
or 

=  nn, 


N-\-i' 


al 


N-\-i' 


N-\-  I 


and  hence 


.TC' 


Fig.  357. 


P=n'EIj,{N-\-i)\ 


As  before,  the  least  value  of  P  corresponds  to  «  =  i,  and 

P=EIj,{N-\-ir 

is  the  least  force  which  will  bend  the  column  laterally. 

Hence,  the  strength  of  the  column  is  increased  in  the  ratio 
of  4,  9,  16,  etc.,  by  causing  it  to  pass  through  points  which  divide 
its  length  into  2,  3,  4,  etc.,  equal  parts,  respectively. 

Cor,  2.  The  value  of  b  may  be  approximately  determined 
as  follows  : 

Let  ds  =  length  of  element  at  M. 

Let  0  =  inclination  to  vertical  of  tangent  at  M. 

Then 

dx 

pressure  upon  ds  =:  P  cos  6  =  P-f 


and  the 


compression  of  ds  = 


_^_.ds  =  ^dx, 


A  being  the  sectional  area  of  the  column. 

Hence,  the  total  diminution  of  the  length  of  the  column 

P 


I 


=    '   ^^" 


EA 


I, 


II 


LONG  COLUMNS  OF   UNIFORM  SECTION.  541 

ii-.gain,  the  length  of  the  cclumn 

=  I  (i-{-  (^-Jj  dx=    Mi  +  a'b'  cos'  ax)dx, 
=    /    ( 1  -j cos  axjax,  approximately, 


Hence,  if  L  is  the  initial  length  of  the  column,  i.e.,  the 
length  before  compression, 


and  consequently 


6' =  2 


EIIL-  I 
'P\     I 


r 

'a: 


M 


Case  2.  Columns  zvith  one  end  fixed  and  the  other  constrained 
to  lie  in  the  same  vertical. 

Assume  that  the  lateral  deviation  is  prevented  x- 
by  means  of  a  horizontal  force  H  at  the  top  of  a  JA 
column.     Then 

-EI^,  =  Py-H{l-x).,    .    .    (I) 

A  particular  solution  of  this  is 

Oz:^Py  -  H{1  -  x\ 

Let^  =^'  -(-  «. 

dy__d\j_ 
■*•  lx~dx'' 


0  V 

Fig.  358. 


§': 


^542  THEORY  OF  STRUCTURES. 

and  eq.  (i)  becomes 


or 


=  —  au. 


.      .     (2^ 


The  solution  of  the  last  equation  is 

y  —  y  =^  u  =  b  sin  {ax  -{•  c),  .     .     .     . 
b  and  c  being  constants  of  integration. 

.' .  y  =  —{I  —  x)  -\-  b  sin  {ax  -\-  c).    .     .     . 


But 


dy 

-p-  =  o    when    x  =^o, 

ax 


and      J  =  o     when    ^  =  o    and  when     x  =■  I. 

H 
.•.  O  =  —  7,  -|-  '''^  COS  ^ ; 

O  =  'p^'V  ^sin^; 
O  =  ^sin  {al-\-c^. 


Hence 


al-\-c  —  o    and    ^/ =  —  tan  ^r  =  tan  a/, 
and  therefore 

which  may  be  written  in  the  form 


(3) 


(4' 


P—  2.045  ;r'' 


EI 


(5) 


LONG   COLUMNS  OF   UNLFORM   SECTION. 
It  is  sufficiently  approximate  to  write 

Ef 


P  ~  2W-' 


/' 


543 


(6) 


0  y 

Fig.  359. 


Case  3.  Columns  zvith  one  end  fixed  and  the  other  free. 

A  rigid  arm  AB  is  connected  with  the  x 

free  end  ^   of   ^  column,  and  a  vertical   b    a  I 

force /^applied  at  B  bends  the  column   j      ^  '^ 

laterally,  until  its  axis  assumes  the  curved   "p 
form  OMA. 

Let  AB  =  q,  AC  =  p,  and  let  /  be  the 
leii^^th  of  the  column,  —  OC,  nearly. 

The  inclination  of  AB  to  the  horizon 
is  so  small  that  the  difference  in  length 
between  AB  and  its  horizontal  projection 
may  be  disregarded.  The  moment  equa- 
tion at  any  point  AI(x,jy)  is 

£/^^.  =  ^^(/> +?-/), 

or 

dy 
Multiplying  each  side  by  2--  and  integrating, 

d  being  a  constant  of  integration. 

dy 
But  ^-  =  o  when  y  =  o,  and  hence  ^  =  o. 


(I) 


•       •       t       • 


(2) 


or 


dy 


V2{p  +  g)y-y 


adx. 


S44 

Integrating, 


THEORY  OF  STRUCTURES. 


COS  -  ^—~~ — -  =ax-\-c, 


c  being  a  constant  of  integration,  or 

But  y  ■=■0  when  ;jr  =  o,  and  hence  c  =.q. 

pJ^q-y 


p^q 


=  COS  ax. 


Also,  y  =  p  when  x  =■  I. 


(3) 


(4) 


.'.  — ^ —  =  COSrt/. 

/  +  !/ 


(5) 


If  ^  is  very  small  or  «//,  the  term  - ■    .        may  be  disregarded, 

and  then 

o  =  COS  al. 


.'.  al 


''~2  =  ^yE'r  •••••• 


(6) 


n  being  a  whole  odd  number. 

The  least  value  of /'corresponds  tow  =  i,  and  the  minimum 
pressure  which  will  cause  the  column  to  bend  laterally  is 


•    (/ 


SI; 


Cor.  I.  By  eq.  (5)  the  deviation  of  the  top  of  the  column 
from  the  vertical  is 


AC=p=q 


I  —  cos  a/ 
cos  al 


(8) 


LONG  COLUMNS  OF  UNIFORM   SECTION. 


545 


Cor.  2.  Let  the  force  applied  at  B  be  oblique  and  let  its 
vertical  and  horizontal  components  be  P  and  H,  respectively. 
The  moment  equation  now  becomes 

EIjl-^=P{p-^q-y)^H{l-x).     ...     (9) 


A  particular  solution  of  this  is 

Let  jj/  =  J/'  -(-  ti. 
Substituting  in  eq.  (9), 


(10) 


or 


d*u 


i  =  —  a  u. 


(II) 


dx 
The  solution  of  this  equation  is 

2t  =  6  sin  {ax  -\-c)  z=y  —  y\ 
^and  c  being  constants  of  integration. 

.'.y=p-^q^p{l-x)^bsm{ax-\-c)..     .    (12) 

dy 
When  x  =  o,y  and  -j-  are  each  =  o ;  and  when  x=zl,y  =  q. 


Hence, 


IT 

o=-p-Vq-\-  -p/+3sin  c\ 


P 


H 


0=  —  -p-\-  ab  cos  c ; 


O  =  P -\-  b  sxn  {al -\-  c) ', 


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Photogi^phic 

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546 


THEORY  OF  STRUCTURES. 


three  equations  giving  b,  c,  and  p,  and  therefore  fully  deter- 
mining y. 

Case  4.  Column  with  both  cuds  fixed. 

Let  i-i  be  the  end  moment  of  fixture.     Then 

X 

d'v 

iix 
or 

'^^=-a'yJ^a'b  =  a\b-y),     .    .    (i) 

1/  ,       M 

where  b  —  p. 

Kio.  300.  Multiplying  each  side  of  the  equation  by  2,-  and 


integrating, 


i^fj=a\2by-f)^d, 


d  being  a  conr-*:ant  of  integration. 

But  '^'-  =  o  when  y  —  o,  and  hence  d  =  0. 
dx 


...(|)=.x%-/)/. 


.     .     (2) 


or 


dy 


V2by  —  y 


^  =  adx. 


Integrating, 


CCS' 


or 


b 

b  —  y 


—  ax  -\-c, 


=  cos  {ax  -f-  c), 


being  a  constant  of  integration. 


(3) 


LONG  COLUMNS  OF   UNIFORM  SECTION, 
But  y  —o  when  x  =  o  and  when  x  =  I.     Hence 
I  =  cos  c  and      I  =  cos  {al-\-  c). 
and  therefore     t  =  o         and     al  =  inn, 
It  being  a  whole  number.     Hence, 


54; 


V^^Z  "'"''' 


or 


P^ie  ./^I 


it^ 


(4) 


The  least  value  of  P  corresponds  to  w  =  i,  and  the  mini- 
mum thrust  which  will  cause  the  column  to  bend  laterally  is 


P  =  ^EI 


71 


(5) 


17.  Remarks. — From  the  preceding  it  appears  that  the 
maximum  theoretical  compressive  strength  of  a  column  per 
unit  of  area  may  be  expressed  in  the  form 


/  = 


r  n' 


.n 


^^A   T'  ~  ^^^'^'ji't 


k  i)cing  the  radius  of  gyration,  and  A  a  coefficient  whose  value 
i-  I,  2,  ^,  or  4,  according  as  the  column  has  tzvo  hinged  tfii/s, 
one  end  fixed  and  the  other  guided  in  the  direction  of  thrust,  one 
cud  fixed  and  the  other  free,  or  two  fixed  ends. 

This  formula  is  easy  of  application,  but  Hodgkinson's 
experiments  show  that  the  value  of  P  as  derived  therefrom  is 
DO  large.  This  may  be  partly  due  to  the  assumption  that  the 
lasticity  of  the  material  is  perfect. 

The  factors  of  safety  to  be  used  with  this  formula  vary 
from  4  to  8  for  iron  and  steel  and  from  4  to  15  for  timber. 

riie  objection  to  the  use  of  flat  bars  as  compression  mem- 
bers has  sometimes  been  overestimated. 

Consider,  e.g.,  the  case  of  a  flat  bar  hinged  at  both  ends. 


■i  ! 
■i  i 


''•'A 


4> 


'•'•  11'  n 


8 


548 


T.IEORY  OF  STRUCTURES. 


Let  the  coefficient  of  elasticity  of  the  material  be  25,000,- 
000  lbs. 

Let  the  working  stress  per  square  inch  be  8000  lbs. 

The  bar  will  not  bend  lat'^rally  under  pressure  so  long  as 

the  unit  stress  <  Ek'j^,  and 


n-d* 


8000  <  2500(xxx)-  ji  ,    or 


2  <  507. 


Hence,  the  length  of  a  flat  bar  in  compression  seems  to  be 
comparatively  limited.  If,  however,  both  ends  are  securely 
fixed,  the  strength  is  quadrupled  and  the  admissible  length  of 
bar  is  doubled,  while  it  may  be  still  further  increased  by  fixine; 
the  bar  at  intermediate  points  as  indicated  in  Corollary  i,  page 
540.  This  shows  the  marked  advantage  to  be  gained  by  rivet- 
ing together  the  diagonals  of  lattice-girders  at  the  points  where 
they  cross  each  other. 

P 

The  value  of  /=  -j  (Art.  15)  must  not  exceed  the  clastic 

limit.  It  is  difficult  to  define  with  any  degree  of  accuracy  the 
elastic  limit  of  cast-iron  and  timber.  It  is  claimed,  indeed,  that 
the  latter  has  no  elastic  limit,  properly  so  called,  but  that  a 
permanent  set  is  produced  by  every  elastic  change  of  form.  It 
may  be  assumed,  however,  that  the  elasticity  of  these  materials 
is  practically  unaffected  so  long  as  they  are  not  loaded  to  more 
than  one  half  of  the  ultimate  crushing  load. 
Hence,  taking 


E 
E 
E 
E 
E 


29,000,000  lbs.     and    /  =  20,000  lbs.  for  wrought-iron, 
29,000,000   "  "      /  z=  33,600   "      "    soft  steel, 

29,000,000   "  "      /  =  56,000  '•      "    hard  steel. 


=  17,000,000 
—     1,500,000 


/  =  40,000   "      "    cast-iron, 
/  =    3,600   "      "    dry  timber. 


/  / 

the  pillars  will  not  bend  laterally  unless  the  ratio  of  - ,  or  — 


LONG  COLUMNS  OF  UNIFORM  SECTION. 


549 


yd  being  the  shortest  side  of  a  rectangular  section  and  r  the 
radius  of  a  circular  section)  exceeds  the  values  given  in  the 
following  table : 


Material.  Value  of  -j. 

a 

Wruught-iron 34.5 

Soft  steel 26.6 

Hard  steel 20.3 

Cast-iron 18.7 

Dry  timber 18.5 

Wrought-iron 48.8 

Soft  steel 37.7 

Hard  steel 28.8 

Cast-iron 26.4 

Dry  timber 26.1 

Wrought-iron 17.2 

Soft  steel 13.3 

Hard  steel 10. 1 

Cast-iron 9,3 

Dry  timber g.2 

Wrought-iron 69 

Soft  steel 53.3 

Hard  steel 40.7 

Cast-iron 374 

Dry  timber 37 


Value  of  — . 


Formula. 


29. 
24 
J7 
10. 
16 
42 
34. 
25- 

22. 
22. 

14 
12 

8. 
8. 
8 

59- 
48 

35 
32 
32 


^  -  A        *^    I' 


P  ifl 


/=3— ? 


Baker  has  deduced  by  experiment  the  following  formulae 
for  tlie  strength  of  wrought-iron  and  steel  pillars  of  from  10  to 
30  diameters  in  length  and  with  fixed  ends,  the  tensile  strength 
of  the  metals  ranging  from  20  to  60  tons  (2240  lbs.)  per  square 
inch  : 

Let  t  be  the  tensile  .strength  of  the  iron  or  steel,  and  H  the 
ratio  of  length  to  diameter. 

Then  the  ultimate  compressive  resistance,  in  pounds  per 
square  inch, 


for  solid  round  pillars 

for  thin  tubes 

for  tubes  with  stiffening  ribs 

for  girder  sections 


(.4  -.oo6//)(/+  18); 
(.44  -  .oo4/^)(^ -f  18); 
(.44-.oo2//)(^+  18); 
(.4    -  .oo4//)(/+  18). 


ji! 


550 


THEOIiY  OF  STRUCTURES. 


18.  Weyrauch's  Theory  of  the  Resistance  to  Buckling. 
— In  order  to  make  allowance  for  buckling,  Weyrauch  piu- 
poses  the  two  following  methods  : 

Method  I.  Let  F^  be  the  necessary  sectional  area,  and  h. 
the  admissible  nnit  stress  for  a  strut  subjected  to  loads  vary- 
ing from  a  maximum  compression  /j,  to  a  minimum  com- 
pression /),. 

Let  F'  be  the  necessary  sectional  area,  and  b'  the  admissible 
unit  stress  for  a  strut  subjected  to  loads  which  vary  between  a 
given  maximum  tension  and  a  given  maximum  compression. 
B'  being  the  numerically  absolute  maximum  load,  and  B"  tlic 
maximum  load  of  the  opposite  kind. 

According  to  Art.  7,  Chap.  Ill,  if  there  is  no  tendency  to 


buckling, 


B. 


.'(1  +  .4) 


(0 


and 


F'  -  ^'  - 

^  -  b'  - 


B' 


I 

v'\\ 


JV  \*     *      *     *      *      * 


(2) 


If  there  is  a  tendency  to  buckling,  let  /  be  the  length  of 
the  strut,  F  its  required  sectional  area,  and  T  the  mean  unit 
stress  at  the  moment  of  buckling. 

Then,  according  to  the  theory  of  long  struts, 


^^      EI        EI 


(3) 


6  being  a  coefificient  depending  upon  the  method  adopted  foi 
securing  the  ends,  E  the  coefificient  of  elasticity,  and  /  tlu 
least  moment  of  inertia  of  the  section. 

Also,  let  /  be  the  statical  compressive  strength  of  the  ma- 
terial of  the  strut,  and  take  t  =  ^T.     Then 


-  1  -lE^ 
^-  T~  6  EI' 


FP_ 
o7' 


(4) 


KESISTANCE    TO  BUCKLING— WE  YRAUCir  S    THEORY.     55 1 


where  <t  = 


8E 


(5) 


If  the  strut  under  a  pressure  /?  were  not  liable  to  buckling, 
it  would  be  subjected  to  a  direct  thrust  only.     The  required 

/; 

sectional  area  of  the  strut  would  then  be  -   ,  <ind  the  unit  stress 

/> 

for  an  area  F  would  be  -., . 

/' 

If  the  strut  under  the  pressure  B  is  liable  to  buckling;,  its 
required  sectional  area  will  be  -— ,  since   T  is  the  mean   unit 

stress  at  the  moment  of  buckling.     Let  x  be  the  unit  stress  iit 
the  moment  of  buckling,  for  the  area  F. 

Assuming  that  the  unit  stresses  in  the  two  cases  are  in  the 
same  ratio  as  the  required  sectional  areas,  ihen 

B      B     B 

A*       •       . •    • •         

Fit 


X  = 


B_l 


B 


(6) 


The  force  which,  when  uniformly  distributed  over  the  area 
F,  will  produce  this  stress,  is  Fx  =  /</>'. 

Hence,  allowance  may  be  made  for  buckling  by  substitut- 
\\\g  for  the  compressive  forces  in  equations  (i)  and  (2),  their 
values  multiplied  by  }x.     Thus,  equation  (i)  becomes 


«i'!. 
<,i1 


txB, 


nB, 


and  equation  (2)  becomes 

^B' 


-/>  -  =  f^f,,     (7) 


^''('-''''«y 


7TT-,  if  B'  is  a  compression,         (8) 


: 


I       I 


.1 


;  i 


552 

and 


B' 

b' 


THEORY  OF  STRUCTURES. 


B' 


,1  ,^B"\ 


— -d77\,  if  B"  is  a  compression.        (9) 


If  yw  <  I,  equations  (i)  and  (2)  give  larger  sectional  areas 
than  equations  (7),  (8),  and  (9),  so  that  the  latter  are  to  be  ap- 
plied only  when  ;<  >  i. 

Method  II.  General  formula;  applicable  to  all  values  of  \x 
may  be  obtained  by  following  the  same  line  of  reasoning  as 
that  adopted  in  the  proof  of  Gordon's  formula.  It  is  there 
assumed  that  the  total  unit  stress  in  the  most  strained  fibre  is 

pA\-\- a  r-,j,Pt  being  the  stress  due  to  direct  compression, and 

p^a  jj  that  due  to  the  bending  action. 

So,  instead  of  employing  equations  (i)  and  (2)  when  /<  <  i, 
and  equations  (7),  (8),  and  (9)  when  /<  >  i,  formulae  including 
a/l  cases  may  be  obtained  by  substituting  for  the  compressive 
forces  in  equations  (i)  and  (2)  their  values  multiplied  by  i  -j-/<. 

Thus,  equation  (i)  becomes 


F  = 


ii+M)B, 


il+M)J'^, (10) 


and  equation  (2)  becomes 

(I  4-  ^^)B" 


F  = 


v'[i 


m 


B' 


(i+A^)^ 


) 


,  if  B'  is  a  compression,    (11) 


or 


F=: 


v'[i  - 


B' 


— /.I     \p/\'  'f  ^"  's  a  compression.  (12) 
m' Ki ) 


B' 


Equation?  (7),  (8),  (9),  respectively,  give  larger  values  of  F 
than  the  corresponding  equations  (10),  (11),  and  (12). 


RESISTAXCE    TO  BUCKLING— WEYRAUCH' S   THEORY.    553 


Dn. 


(9) 


ional  areas 
i  to  be  up- 
values of  )X 
sasoning  as 
It  is  there 
ined  fibre  is 

)ression,  and 


when  /i  <  I, 

ilae  including 

;  compressive 

edby  !+/*• 


.    .    .    (10) 


Iression,   (n) 

)ression.  (I2) 
values  of  f 

1 1 2). 


Note, — For  wrought-iron  bars  it  may  be  assumed,  as  in  Arts. 
5,  6,  Chap.  Ill,  that  x\  =  v'  =  700  k.  per  sq.  cm.,  and  w,  =  ;«' 

The  value  of  o  is  given  by  formula  (5),  but  is  unreliable,  and 
varies  in  practice  from  io,CXX)  to  36,cxx)  for  struts  with  fixed 
ends. 

When  the  ends  are  fixed,  d  =  4^-',  according  to  theory. 
Hence, 

a  =  4-'  J. 

Therefore,  \{  E  =■  2,000,000  k.  per  sq.  cm.,  and  t  =  3300  k. 
per  sq.  cm.,  6  =  23,926,  or  in  round  numbers  23,900;  24,000 
is  the  value  usually  adopted  by  Wcyrauch. 

Example. — The  load  upon  a  wrought-iron  column  360  cm. 
long  varies  between  a  compression  of  50,000  k.  and  a  compres- 
sion of  25,000  k.  Calculate  the  sectional  area  of  the  column, 
assuming  it  to  ht  first  solid  and  j^rt?;/^/ hollow,  allowance  being 
made  for  buckling. 

First.  By  eq.  (i), 


^.= 


50000 


700(1 +ix  Mm) 


400      , 

/ 


r  being  the  radius  of  the  section. 


Also,    /  = 


nr^ 


"  /  ~  r'  ~  50' 


Hence,  by  eq.  (4), 

360  X  360 


fji  = —  X  —  =  1. 188. 

24000  50 

Thus  ^  >  I,  and  by  eq.  (7)  the  required  sectional  area  is 

F,  X  1. 188  =  A^^  X  1. 188  =  67.9  sq.  cm. 

Second.     F,  =  ^^=  n{r*  -  r,'), 

>",  being  the  external  and  r,  the  internal  radius  of  the  section. 


554 


THEORY  OF  STRUCTURES, 


Let  /",  =  9  cm.  and  r,  =  7.92  cm.     Then 


'rl^'  -  r^)  -  57.43  sq.  cm. 


Also,     /  = 


71 


(' 


K) 


I       r;'  +  r,'       143-7^64 


Hence,  by  eq.  (4), 


h  = 


360  X  3^XD 


—  X 


4000  1437-64 


=  .15. 


cncy 


Thus,  in  the  latter  case,  since  /<  <  i,  there  is  no  tend 
to  buckling. 

If  the  area  is  determined  by  equation  (10),  its  value  become^ 
I.I 5  X  A 2 a  — 65  sq.  cm. 

19.  Flexure  of  Columns. — In  Art.  16  the  moment  equa- 
tion has  been  expressed  in  the  form 

and  this  is  sufficiently  accurate  if  the  deviation  of  the  axis  of  the 

(dyV 
strut  from  the  vertical  is  so  small  that  ( -f-1   may  be  neglected 

without  sensible  error. 

The  more  correct  equation  is 


P 


p  being  the  radius  of  curvature. 

Consider,  e.g.,  the  strut  in  Art.  16,  Case  I.     Then 


P  \      dd      dd    ,     ^ 

•^  EI'         li       ds       dy 


no  tendency 
alue  beconur 
[lomcnt  cqua- 


FLEXURE   OF  COLUMNS. 


555 


'^bciny  the  inclination  of  the  tangent  at  J/ to  the  axis  of  a", 
and  ds  an  element  of  the  bent  strut  at  J/. 


lntc<;rating, 


.'.  —  d'ydy  =  sin  f^Jff. 


d'y 

=  cos  0  —  cos  6„ , 

2  »' 


(I) 


K,^  being  the  value  of  ft  at  a  strut  end. 

Let  sin  —  =  u     and     sin  -  =  u  sin  0.     Then 

2  2 


«y 


=  2yM'(l  —  sin''0), 


or 


2/A 

y  =  —  cos(p (2) 


Let  Fbe  the  maximum  deviation  of  the  axis  of  the  strut 
from  the  vertical,  i.e.,  the  value  of  y  when  ft  =  o  or  0  =  o. 
Then 


F^ 


2h 

a 


sin  4 


a 


(3) 


Again, 


ds  —  pdft  =  —■ 


de 


^  \  \  —  jji'  s\n'  0 
Hence,  if  /be  the  length  of  the  strut, 

d(t'i 


2      /*"  (10  2 

«  /       \  I-  fu'  sin''  0       «      ^ 


(4) 


F„(0)  being  an  elliptic  integral  of  the  Jirst  kind. 

Let  P'  be  the  /m.y/  thrust  which  will  make  the  strut  bend. 
As  shown  in  Art.  16, 


I  im 


5 $6  THEORY  OF  STRUCTURES. 

and,  by  eq.  (4),  the  corresponding  value  of  the  modulus  /<  is 
given  by 

PM)  =  \ (5) 

Let  the  actual  thrust  on  the  strut  be 

P=n'F (6) 

«*  being  a  coefficient  >  unity. 

The  corresponding  value  of  the  modulus  is  given  by 

j:,,^.       I     I~P       I  n 

FA<P)  =  2  Y  -^j  =  ^na  =  n- {-jj 

By  reference  to  Legendre's  Tables  it  is  found  that  a  large 
increase  in  the  value  of  yu,  i.e.,  of  sin  -"  or  6^,  is  necessary  in 
order  to  produce  even  a  small  increase  in  the  value  of  h\{(p) 
and  therefore  of  ;/i  =  \/  'n't  )•     Hence,  as  soon  as  the  thrust 

P  exceeds  the  least  thrust  which  will  bend  the  column,  viz.. 
P',  6^  rapidly  increases. 

The  total  maximum  intensity  of  stress  in  the  skin  of  the 
strut  at  the  most  deflected  point 

P       Mz      P  ■    PYz       ^  ,   22   .    6,     l~ 

=  :^-  +  -7-  =  z  +  -7-=^+T^'"2V/^'  •  ^«' 

z  being  the  distance  of  the  skin  from  the  neutral  axis,  and  / 

P 

being  equal  to  -j. 

The  last  term  of  this  equation  includes  the  product  fE. 

a 

which  is  very  large,  and  also  the  factor  sin  -  ,  which  increases 

with  6^  so  that  the  ultimate  strength  of  the  material  is  rapidly 
approached,  and,  in  fact,  rupture  usually  takes  place  before  the 
column  has  assumed  the  position  of  equilibrium  defined  by  the 
slope  6^  at  the  ends. 


FLEXURE  OF  COLUMNS. 


557 


If  there  were  no  limit  to  the  flexure,  the  column  would 
tako  its  position  of  equilibrium  only  after  a  number  of  oscilla- 
tions about  this  position,  and  the  maximum  stress  in  the 
material  would  be  necessarily  greater  than  that  given  by 
eq.  (8). 

Again, 

,  ,  _  I  (i  -  2/i'  sin'  0y0 

ax  =  (IS  cos  6  = ^ — ■ -^, 

rt     Vi  —  pi'  sin"  0 

Let  X  be  the  vertical  distance  between  the  strut  ends.    Then 


w 

2     /*'  I  —  2fA*  sin*  ff> 
~^J,     V\  -  A'sin''0''^ 


;-|^3^(i-;.'sin>K0-y^>^_^t,.„,^j 


n 


=  ^!2£,(0j-/v(0)[; 


£^(0)  being  an  elliptic  integral  of  the  second  kind. 
Hence,  the  diminution  in  the  length  of  the  strut 


=  L-X=l\F,{<t>)-E,{<P)\, 


20.  Flexure  of  Columns  (Findlay). — In  a  paper  on  the 
flexure  of  columns  read  before  the  Canadian  Society  of  Civil 
Engineers  (Vol.  IV,  Part  I),  Findlay  expresses  the  moment 
equation  in  the  form 


0,  and    ,—  being  the  values  of  p  and  — p-  when  M  =  o. 

(tx  ^  ax 


I 


m 


M 


558 


TnEORY  OF  STRUCTURES. 


Hinged  Ends. — It  is  assiinud  that  the  line  of  action  uf  the 
thrust  P  is  at  a  distance  d  from  the  axis  of  the  strut.    Then 


--'(i^S")=-o.+<o^^. 


(2) 


or 


where  c^  =^  v,, ,  />  —  total   stress  at   the  distance  c  from  tii-j 
/:/ 

neutral  axis,  and/—  stress  due  to  direct  thrust  (=;  —1,  so  ihat 

the  stress  due  to  /fending  =  /»—/. 

It  is  also  assiuncd  that  the  form  of  the  axis  of  the  column 
before  it  is  acted  upon  by  the  thrust  /',  is  a  curve  of  sines 
defined  by  the  equation 


TtX 

J',  =  J  cos  ~,     .     . 


(4' 


the  origin  being  half-way  between  the  ends  of  the  strut,  and  J 
being  the  maximum  initial  deviation  of  the  axis  from  the  ver- 
tical, i.e.,  the  value  of  j'„  when  .r  =  o. 


dx' 


An*         itx 

-F  '^^^  T' 


and  hence,  by  eq.  (3), 


dx 


-.r  =  -a\y  +  d)  —  Aj,  cos  -j. 


.    (5) 


A  solution  of  this  equation  is 


cosa.r 
cos  - 


cos 


TTX 


{6^ 


I  — 


n' 


FLEXURE    OF  COLUMNS. 


559 


Now  —  is  alwavs  small  for  such  values  of  /  as  would  con- 
-titute  a  safe  working  load,  and  therefore 


al 
cos  -   =  I 

2 


,5 /J 

a  I 


-,  approximately, 


i'T 


or 


tliat  eq.  (6)  becomes 
</ —  </ COS  a,i\i ^1     4"  J  cos  -7- (I r)    > 


• -|-  «  =  «  cos  a.v\\  ~\ ^j  +  A  cos  — r  I  i  -| ,- j,  approx.    (7) 

Let  Fbe  the  maximum  value  of  j'.  i.e.,  the  value  oi y  when 
.1-  z^  0.     Then 


y =<+-)+ '^■ 


(8i 


Hence,  by  eq.  (3),  the  total  maximum  intensity  of  stress 


where     /;  =  -   -  J — -I      and     c 
s  \o         It  I 


d-^  A 


Eq.  (9)  is  a  quadratic  from  which /may  be  found  in  terms 
if/.  As  a  first  approximation,  />  may  be  siibstitutetl  for  /  in 
lie  last  term  of  the  portion  within  brackets,  tlie  error  beini;  in 
he  direction  of  safety. 

Fixed  Ends.— hct  J/,  be  the  moment  of  fixture. 

Eq.  (3)  now  becomes 


=  -a{j'  +  ^+^)-     (10) 


5 


III 


560 


THEORY  OF  STRUCTURES. 


Assuming  again  that  the  initial  form  of  the  axis  is  a  curve 
of  sines,  the  solution  of  the  last  equation  is 


cos 


nx 


Initially, 


(n.i 


cos 


I  — 


y^  =  A  cos  -y, 

dy  dy  I  I  ' 

and  -7-  is  equal  to  ~  when  ;tr  =  —  or  = . 

dx        ^  dx  22 


Hence, 


-.7=-K 


.    al 


71 


PI  al 

cos  — 

2 


a'P 


n 


or 


<^+f  = 


(12) 


Again,  the  value  of  j  at  the  point  a-  =  o  is 
Y 


PI    8 


(13' 


Also,  '^i Pi,  Pi  are  the  total  maximum  intensities  of  stress  at 
the  end  and  at  the  most  deflected  point,  then 


^^--^'('^+^)=etc.,    ....     (.4^ 


and 


A-/ 
zE  ' 


=  -AY+ 


-+?■) 


etc. ;    . 


(15* 


two  equations  from  which  /  may  be  found  as  before. 

The  following  conclusions  are  drawn  from  the  above  inves- 
tigation : 

First.  The  actual  strength  of  a  column  depends  partly  upon 


FLEXURE   OF  COLUMNS. 


561 


i  IS  a  curve 


known  facts  as  to  dimensions,  material,  etc.,  and  partly  upon 
accidental  circumstances. 

Second.  Experiments  upon  the  crippling  or  destruction  of 
columns  cannot  be  expected  to  give  coherent  results  when 
applied  to  the  determination  of  the  constants  in  such  an  equa- 
tion as  No.  (9). 

Third.  It  is  a  question  whether  p  should  be  made  the 
dastic  liuiit  of  the  material  and  the  working  load  a  definite 
fraction  of  the  corresponding  value  of  /  derived  from  eq.  (9), 
or  whether/  should  be  the  allowable  skin  working  stress,  and 
the  working  stress,/  be  found  by  means  of  the  same  equation. 
The  former  seeVns  to  be  the  more  logical  assumption. 

Fourth.  It  would  appear  that  the  strength  of  hinged  col- 
umns is  likely  to  be  much  more  variable  than  the  strength  of 
columns  with  fixed  ends,  as  it  depends  upon  two  variable 
elements  d  and  J,  while  the  end  fixture  eliminates  d. 


Note, — The  Tables  on  the  following  page  give  the  numerical 
values  of  elliptic  integrals  of  the  first  and  second  kind,  and  are 
useful  in  applying  the  results  of  Art.  18. 


••t^=. 


rir 


■■'p 


(.i- 


fi 


[U    1 


1 1 


ri    i 


i 


i1    ! 


562 


THEORY  OF  STRUCTURES. 


FIRST   ELLIPTIC   INTEGRAL 

.    FM). 

♦ 

H  =  0 

ji  =  .1 

fi  =  .J 

M  =  -7. 

M  =  .4      M  =  .5 

H  =  .6 

M  =.7 

1 
H  =.8 

ft  =  .9    »»  =  . 

0° 

0.000 

0.000 

0.000 

0.000 

0.000    0.000 

0.000 

0.000 

0.000 

0.000    0.000 

5° 

0.087 

U.087 

0.087 

0.087 

0.087  ,  0.087 

0,087 

O.0S7 

O.0S7 

0.087    0,087 

10 

0.175 

0.175 

0.175 

0.175 

0.175   0.175 

0.175 

0.175 

0.175 

0.175    0.175 

*s: 

0.262 

0.262 

0.262 

0.262 

0.262    0.263 

0.263 

0.263 

a.  264 

0.264    0.265 

20" 

o-34<) 

0.349 

0.349 

0.350 

0.3501  0.351  j  0.352 

0.353 

0.354 

0.355    0.350 

25 

0.436 

0.436    0.437 

O.43S 

1 
0.439   0.440 , 0.441 

0.443 

0.445 

0.448    0.451 

30 

0.524 

0.524 

0.525 

0.526 

0.527 1  0.529 

0.532 

0.536 

0.539 

0.544    0  549 

35 

0.6  n 

0.61 1 

0.612 

0.614 

0.617  1  0.620 

0.624 

0.630 

0.636 

0.644    0.653 

4°: 

0.698 

0.699 

0.700  1  0.703 

0.707    0,712 

0.718 

0  727      0.736 

0.748    0.763 

45^ 

0.785 

0.786 

0.789 

0.792 

0.798    0.804 

0.814 

0.826      0.839 

0.858    o.cSi 

SO 

0.873 

0.874 

0.877 

0.882 

0.889   0.898 

0.911 

O.92S 

0.947 

0.974     lOII 

15" 

0.960 

0.961 

0.965 

0.972 

0.981    0.993 

1. 010      1.034 

I  060 

1.099    1.1:4 

1.047 

1.049     1.054 

1.062 

1.074     I.<J90 

1. 112      1. 142 

1. 178 

1.233   1..U7 

65^ 

I-I34 

I-I37  ;  1. 143 

I.I53 

1. 168    1.187 

I.2!5       1.254 

1.302 

1.377     1-5"" 

70^ 

1.222 

1.224 

1.232 

1.244 

1.262  1  1.285 

1.320 

1-370 

1-431 
1.566 

1-534    1-735 

% 

1.309 

1. 312    1. 321 

1.336 

1-357     1.385 

1.426 

1.4S8 

1.703    2.02S 

1.396 

1.400  1.410  1  1.427 

1.452     1.485 

1-534 

1. 60S 

1-705 

1.S85  1  2.4V) 

«S^ 

1.484 

1.487   1.499 

1-519 

1.547  j  1.585 

1.643 

1. 731 

1.84S 

2.077    3.131 

90^ 

I-57I 

1.575  1.588 

1.610 

1.643     1.686 

1.752 

1.854 

1.993 

2.27.1.       " 

i 

SECON 

D   ELLIPTIC    INTEGR.AL.    E^{<p). 

* 

f*  =  0 

M  =  .1 

M  =  .a 

M  =  .3 

M  =  .4 

ft  =^-5 

H  =  .6 

ft  =  -7 

M  =  .8 

ft  =  .9 

ft  =  I 

0° 

0.000 

0.000 

0.000 

0.000 

0.000 

0.000 

0.000 

0.000 

0.000 

0.000 

coco 

5° 

0.087 ;  0.087 

0.087 

0.087 

0.087 

0.087 

0.087 

0.087 

0.087 

0.087 

0.0S7 

10" 

0.175   0.175 

0.174 

0.174 

0.174 

0.174 

0.174 

0.174 

0.174 

0.174 

0.174 

i.S'' 

0.262  0.262 

0.262 

0.262 

0.261 

0.261 

0.261 

0.260 

0.260 

0.259 

0.259 

20" 

0.349  0.349 

0.349 

0.348 

0.348      0.347 

0.347 

0.346 

0.345 

0.343 

0.342 

25' 

0.436  0.436 

0.436 

0.435 

0.434  I  0.433 

0431 

0.430 

0.428     0.425 

"-423 

30 

0.524  0.523 

0.523 

0.521 

0.520  i  0.518 

0.515 

0.512 

0.509     0.505 

o.:o() 

35" 

0.611   0.610 

0. 609 

0.607 

0.605  1  0.602 

0.598      0.593 

0.588   :   0.581 

0.5:4 

40 

0.698    tj.698 

o.6<)6 

0.693 

0.690  1  o.()85 

0.679  1  0.672    <j.664  I  0.654 

o.('43 

45' 

0.785    0.7S5 

0.7S2 

0.779 

0.773  !  0.767 

0.759 

0.748 

0.737   j  0.723 

0.707 

.50 

0.873    0.872 

0.869 

0.864 

1 
0857    0.848 

0.837 

0.823 

U.808      0.789 

o.7()6 

ir 

0.960    0.959 

0.955 

0.948 

0.939 

0.928 

0.914 

O.S95 

0.875       0.850 

0.819 

1.047 

1.046 

1. 041 

1.032 

1.021 

1.008 

0.989 

0.965 

0.940      0.907 

O.fl.'l 

65" 

1. 134 

1. 132 

1. 126 

1. 116 

1. 103 

1.086 

1.063 

1.033 

1. 00 1     0.960    o.()u6 

70 

1.222 

1.219 

1. 212 

1.200 

1. 184 
1.264 

1.163 

1.135 

1.099 

1.060  '  1.008    0.941 

t- 

1.309 

1.306 

1.297 

1.283 

1.240 

1.207 

1.163 

1. 117     1. 053.0.066 

1.396 

1.393 

1-383 

1.367 

1.344 

1.316 

1.277       1.227 

1.172     1.095  ;  0.9S; 

85" 

1.484 

1.480 

1.468 

1.450 

1.424 

1.392 

1.347    ;    I.2S9 

1.225     1.135    O.qiit) 

90° 

1. 571 

1.566 

1-554 

1-533 

1.504 

1.467 

I.4I7        1.351 

1.278    I.i73j 

1 

1. 01)0 

B     «*  =  -9 


EXAMPLES. 


563 


0  1  0.000    ( 

).000 

7    < 

3.087 

j.iiS; 

5 

0.175 

0.175 

4 

0.264 

0.265 

4 

0.355 

o.35'i 

—  ;- 



15 

0.448 

0.451 

59 

0.544 

0  M'J 

3^ 

0.644 

o.6;3 

36 

0.748 

o.7'''3 

39 

0.85S 

O.SOI 

47 

0.974 

I  oil 

60 

1.099 

i.'M 

-8 

1.233 

1.317 

i02 

1.377 

1.51)6 

*3i 

'  1-534 

1.735 





' 

566 

1.703 

2.023 

705 

1.885 

1  2.43<' 

843 

2.077 

'  3-131 

993 

2.27;- 

CO 

I428  0,425 
I509 !  0.505 
Uss  ,0.581 


0.423 

O.fil'l 

o.;:4 
0.643 
0.70; 


117 
I172 
I225 
I278 


1.053    0.066 

i.oc)5  1°-''*- 
1.135    O'^'* 

1 1. 173 :  ''"^ 


•        EXAMPLES. 

1.  A  Phcenix  column  in  four  segments,  each  weighing  17  lbs.  per 
lineal  yard,  carries  a  load  of  68,000  lbs.  What  is  the  compressive  unit 
stress?  Ans.  10,000  lbs.  per  sq.  in. 

2.  The  sectional  area  of  a  pillar  is  144  sq.  in.,  and  the  pillar  carries 
a  load  of  4000  lbs.  Find  the  normal  and  tangential  intensities  of  stress 
on  a  plane  inclined  at  20'  to  the  axis.  Ans.  3.25  lbs.;  8.93  lbs. 

3.  A  short  hollow  square  column  has  to  support  a  load  of  120,000  lbs., 
the  allowable  stress  being  15,000  ll)s.  per  square  inch.  Find  the  thick- 
ness of  the  metal,  an  external  side  of  tlie  column  being  6  in. 

Atis.  .36  in. 

4.  A  solid  cast-iron  pillar  9  ft.  in  height  and  4  in.  in  diameter  sup- 
ports a  load  of  55,000  lbs.  Find  the  normal  and  shearing  intensity  of 
stress  per  square  inch  in  a  plane  section  inclined  at  30'  to  the  axis. 

If  the  ends  of  the  pillar  are  flat  and  firmly  bedded,  determine  its 
breaking  weight,  t)oth  by  Hodgkinson's  and  by  Gordon's  formula. 

Ans.  1093I  lbs. ;   1894.375  lbs. ;  141  i  tons  by  H.;  159  tons  by  G. 

5.  A  cylindrical  pillar  6  in.  in  diameter  supports  a  load  of  400  lbs., 
of  which  the  centre  of  gravity  is  f  in.  from  the  axis.  Determine  the 
greatest  and  least  intensities  of  stress  upon  any  transverse  section  of 
the  pillar.  Ans.  25*!  lbs. ;  2j|5  lbs. 

6.  Compare  the  breaking  weights  of  round  cast-iron,  wrought-iron, 
and  mild-steel  pillars  with  flat  and  firmly  bedded  ends,  each  being  9  ft. 
in  length  and  6  in.  in  diameter. 

Ans.  1,250,197  lbs. ;  890,10911)5.;  1,543,572  lbs. 

7.  .\  hollow  cast-iron  pillar  with  an  external  diameter  of  9  in.  is  to 
be  substituted  for  the  solid  pillar  in  the  preceding  question.  Determine 
the  thickness  of  the  metal.  Ans.  \  in. 

8.  Determine  the  breaking  weight  of  a  solid  round  pillar  with  both 
ends  firmly  secured,  10  ft.  in  length  and  2  in.  in  diameter,  (ij  if  of 
cast-iron  ;   (2)  if  of  wrought-iron  ;  (3)  if  of  steel  (miUlj. 

Ans.  25142.8  lbs.;  43516.48  lbs. ;  54,36  lbs. 

9.  A  hollow  cast-iron  pillar  12  ft.  in  height  has  to  support  a  steady 
load  of  33,000  lbs.;  its  internal  diameter  is  5i  in.  Find  the  thickness 
of  the  metal,  the  factor  of  safety  being  6.  Ans.  .28  in. 

10.  A  solid  wrought-iron  pillar  is  to  be  substituted  for  the  pillar  in 
the  preceding  question.     Find  its  diameter.  Ans.  3^  in. 


564 


THEORY  OF  STRUCTURES. 


t'  I 
"Hi 

! 


I 'I' 
I' 


¥ 


11.  What  is  the  breaking  weight  of  a  hollow  cast-iron  pillar  9  ft.  in 
length  and  6  in.  square,  the  metal  being  i  in.  thick  ? 

Ans.  970873.6  lbs. 

12.  Compare  the  breaking  weight  of  a  solid  square  pillar  of  wrouj^lu- 
iron  20  ft.  long  and  6  in.  square  with  that  of  a  solid  rectangular  pillar 
of  the  same  material,  the  section  being  9  in.  by  4  in. 

Ans.  845,217  lbs.;  589,090  lbs. 

13.  Compare  the  breaking  weights,  as  derived  from  Hodgkinsoii's  and 
Gordon's  formulae,  of  a  solid  round  cast-iron  pillar  20  ft.  in  length  ami 
10  in.  in  diameter,  (i)  both  ends  being  securely  fixed ;  (2)  both  ends 
being  imperfectly  fixed. 

Ans. — (i)  951.4  tons  by  H.;  1150.05  tons  by  G. 
(2)  280.6  tons  by  H.;  415  tons  by  G. 

14.  Determine  by  Hodgkinson's  formula  the  diameter  of  a  solid 
wrought-iron  pillar  equal  in  length  and  strength  to  that  in  the  preceding; 
question.  Ans.  7.35  in. 

15.  A  solid  or  hollow  pillar  of  cast-iron,  wrought-iron,  or  mild  stee' 
is  to  be  designed  to  carry  a  steady  load  of  30,000  lbs.  Detei mine  the 
necessary  diameter  in  each  case,  6  being  a  factor  of  safety.  (The  pillar 
is  to  be  12  ft.  high,  and  the  metal  of  the  hollow  pillar  is  to  be  |  in. 
thick.)  Ans. — Solid:       3.42  in.;  3.25  in.;  2.8  in. 

Hollow:  4.5    in.;  4.75  in.;  3.5  in. 

16.  Determine  the  load  in  the  preceding  question  that  will  produce 
a  maximum  stress  of  9000  lbs.  per  square  inch  in  the  solid  steel  pillar. 

17.  A  pillar  of  diameter  D  supports  a  given  load  ;  if  A'^  pillars,  eacii 
of  diameter  d,  are  substituted  for  this  single  pillar,  show  that  r/must  lie 

between  — -.  and  — r. 

A'*  ^V* 

Is  it  more  economical  to  employ  few  or  many  pillars  of  given  height 
to  support  a  given  load  ? 

18.  A  solid  round  pillar  of  mild  steel,  16  ft.  high,  supports  a  steady 
load  of  20,000  lbs.     If  the  factor  of  safety  is  6,  what  is  its  diameter  } 

Ans.  3  in. 

19.  Find  the  diameter  of  each  of  four  pillars  of  the  same  material 
which  may  be  substituted  for  the  single  pillar  in  the  preceding  question. 

Ans.   2.04  in. 

20.  What  is  the  breaking  weight  of  a  cast-iron  stanchion  of  a  ren;ular 
cruciform  section  and  15  ft.  in  height,  the  arms  being  24  in.  by  i  in.  .' 

Ans.  2,811,215  lbs. 

21.  Each  of  the  pillars  supporting  the  lowest  floor  of  a  refinery  is 
6i   ft.  high,  is  of  a   regular  cruciform  section,  and  carries  a  load  of 


EXAMPLES. 


565 


lillar  9  ft.  in 

0873.6  lbs. 
of  wr(jui;lil- 
ngular  pillar 

589,090  lbs. 

ikinsuii's  aiul 
in  length  and 
2)  both  ends 

tons  by  G. 
ns  by  G. 

er   of  a  solid 
the  preceding 
4ns.  7.35  '"• 
1,  or  mild  slee' 
determine  the 
^     (The  (jiUar 
•  is  to  be  \  ill. 
25  in.;  2.8  in. 
5  in.;  3-5  '"■ 
,t  will  produce 
steel  pillar. 

pillars,  eacli 
lihat  d  must  lie 


|f  given  heiL^lit 

Lports  a  s^tcady 

[diameter  ? 
Ans.  3  in. 

[same  material 

Iding  question. 
ins.  2.04  in. 
DH  of  a  rei,'ular 
In.  by  I  ill'  ? 
1,811,215  lli^- 
If  a  refinery  is 
1-ies  a  load  of 


240,000  lbs.;  the  total  length  of  an  arm  is  26  in.     Determine  its  thick- 
ness, the  factor  of  safety  being  10.  Ans.  2.558  in. 

22.  Find  the  breaking  stress  per  square  inch  of  a  4-in.  x  4-in.  solid 
wrought-iron  pillar  for  lengths  of  5,  10,  15,  and  20  ft.,  the  two  ends 
being  absolutely  fixed. 

Ans.  33,488  lbs.;  27,692  lbs.;  21,492  lbs.;  16,363  lbs. 

23.  Find  the  diameter  of  a  wooden  column  20  ft.  long,  to  support  a 
load  of  10,000  lbs.,  10  being  a  factor  of  safety  and  both  ends  of  the 
column  being  absolutely  fixed.  Ans.  8.55  in. 

24.  The  external  and  internal  diameters  of  a  hollow  cast-iron  column 
12  ft.  in  length  are  D  and  \D,  respectively;  the  load  upon  the  column 
is  25,000  lbs.  If  the  factor  of  safety  is  4,  find  D,  (a)  when  both  ends  of 
the  column  are  absolutely  fixed ;  (b)  when  both  ends  are  hinged. 

Ans.  (a)  4.3  in.;  (6)  5.4  in. 

25.  Determine  the  breaking  weight  of  an  oak  pillar  9  ft.  high,  11  in. 
wide,  and  5  in.  thick.  Ans.  138,160  lbs. 

26.  What  weight  will  be  safely  borne  by  a  pillar  of  dry  oak  subject 
to  vibration,  10  ft.  high  and  6  in.  square,  10  being  a  factor  of  safety  ? 

Ans.  9969  lbs. 

27.  The  web  members  of  a  Warren  girder  are  bars  of  rectangular 
section  and  10  ft.  in  length.  One  of  the  bars  has  to  carry  loads  varying 
between  a  steady  maximum  tension  of  20.2  tons  and  a  maximum  tension 
of  40.4  tons,  and  another  to  carry  loads  varying  between  a  maximum 
compression  of  8.7  tons  and  a  maximum  tension  of  14.4  tons.  Find  the 
sectional  area  in  each  case,  allowance  being  made  for  buckling  in  the 
latter. 

28.  Determine  the  sectional  area  of  a  double-tee  strut  which  is  to 
carry  a  load  varying  between  a  maximum  tension  of  80,000  lbs.  and  a 
maximum  compression  of  60,000  lbs.  Each  flange  consists  of  two  6- 
in.  X  6-in.  x  |-in.  angle-irons  riveted  to  a  12-in.  x  fin.  web  plate. 
The  length  of  the  strut  is  to  be  (a)  6  ft.;  (i)  12  ft. 

29.  A  steel  strut  lo  ft.  long  consists  of  two  tees  back  to  back,  each 
4 in.  X4in.  x  ^in..  Taking/ =  60,000  lbs.,  ai  ■=  T,y?,5Tj  (page  526),  and  6 
as  a  factor  of  saf£'..y,  find  the  working  load  {a)  when  ihe  strut  has  two 
pin  ends;  {b)  when  it  bar.  two  fixed  ends.     {E  =  29,000,000  lbs.) 

Also,  find  the  deviation  of  the  axis  of  the  load  from  the  axis  of  the 
strut  so  that  the  maximum  stress  in  the  metal  may  not  exceed  10,000 
lbs.  per  square  inch. 

Ans — (a)  25,585  lbs.;  (b)  52,229  lbs. 

Deviation  =  .55  in,  in  (a)  and  .158  in.  in  (d). 

30.  A  solid  wrought-iron  strut  20  ft.  high  and  4  in.  in  diameter  has 
one  end  fixed  and  the  other  perfectly  free.     Find  the  deviation  of  the 


It 


-,  i- 1      ' 
5    ; 

1 


566 


THEORY  OF  STRUCTURES. 


line  cf  action  of  a  load  of  10,000  lbs.  from  the  axis,  so  that  the  stress 
may  not  exceed  10,000  lbs.  per  square  inch,  E  being  27,000,000  lbs. 

Alls.  .88  in.  if  P  =  ^  ;     1.8  in.  if  Z'  =  ^ EI^~ 
I'  4     /' 

31.  A  hollow  cast-iron  column  with  two  pin  ends  is  24  ft.  hi},'li  and 
has  a  mean  diameter  of  12  in. ;  it  carries  a  load  of  80,00c  lbs.  Find  the 
proper  thickness  of  the  metal,  10  being  a  factor  of  safety.  If  the 
deviation  of  the  line  of  action  of  the  load  from  the  axis  is  1  in.,  find 
the  maximum  stress  per  square  inch  in  the  metal,  E  being  i7,ooo,cxx) 
lbs.  Ans.  1.28  in.;  2236  lbs.  per  sq.  in. 

32.  Find  the  crushing  load  of  a  solid  wrought-iron  pillar  3  in.  in 
diameter,  10  ft.  high,  and  fixed  at  both  ends.  Calculate  the  deviation 
which  will  produce  a  maximum  stress  in  the  metal  of  9000  lbs.  per 
square  inch  under  loads  of  \i.i)  15,000  lbs.;  {J))  30,000  lbs.;  E  being 
29,000,000  lbs.  Ans.   148,775  lbs.    {a)  1.158  in. ;  {b)  .38  in. 

33.  Solve  the  preceding  question  on  the  assumption  that  the  colunui 
has  two  pin  ends.  Ans.  66,218  lbs.  ;  {a)  .985  in.;  (b)  .261  in. 

34.  A  pier  consists  of  A''  rows  of  posts  equidistant  from  each  other, 
A^  being  even;  d  is  the  distance  from  centre  to  centre  of  the  outside 
rows;  W^  is  the  gross  vertical  load  upon  the  pier;  H  is  the  greatest 
horizontal  thrust,  and  acts  upon  the  pier  at  a  hei<,'ht  y  above  the  base. 
Assuming  the  principle  of  a  uniformly  varying  stress,  the  portion  of  Uie 
load  borne  by  the  «-th  row  of  posts  measured  from  the  centre  line  is 

Y  • ■ — -.     Find  the  value  of  the  coefficient  a  in  terms  of  </,  //, 

JV         2  iV  —  I 

^,  and  A^,  and  determine  the  best  value  for  </. 

35.  Prove  that  the  flexural  rigidity  of  a  straight  beam,  of  sectional 

area  A,  under  a  thrust  P  per  unit  of  area,  is  EAi^(  i  —  -pj-  and  that 
beam  will  bend  if  its  length,  when  unstrained,  exceeds 


the 


'Vf -(■-!)• 


AJt^  being  the  moment  of  inertia  of  the  section,  and  £^  the  coefficient  of 
elasticity  of  the  'material. 

36.  Find  the  safe  load  on  a  rolled  tee-iron  strut  6  in.  x  4  in.  x  i  in., 
10  ft.  long,  fixed  at  one  end,  free  at  the  other. 

37.  In  Art.  19,  show  how  equations  (3)  and  (6)  will  be  modified  if  the 

line  of  action  of  P  is  distant  a  +  fi  from  one  end  and  a  —  ft  from  the 

other  end  of  the  column's  axis.     Also,  if  the  coefficient  of  elasticity,  E, 

z 
is  variable  and  equal  to  m  ±  «-  at  a  point  distant  z  from  the  axis,  /■  being 


I  ) 


EXAMPLES. 


567 


at  the  stress 
,000  lbs. 


4  ft.  high  and 
bs.  Find  the 
ifety.  H  the 
s  is  1  ill.,  find 
ing  17,000,000 
s.  per  sq.  in. 

pillar  3  in.  in 
the  deviation 
■  9000  lbs.  per 
lbs.;  E  beini^ 
1.;  (*).38'n. 
hat  the  coUimn 
.;  (b)  .261  in. 
om  each  other, 
of  the  outside 
is  the  greatest 
above  the  base. 
portion  (if  the 
centre  line  is 

terms  of  d,  //. 


the    maximum    value    >jf    ::,   and     vi    and    n    coefficients,   show    that 

n  /{•' 

)■  4-  —  —  must  be  substituted  for_j'  in  eq.  (3). 
Ill  r 

38.  In  one  of  Christie's  experiments  an  angle-bar  2  in.  X2  in.  x  ,'j  in., 

wall  hinged  ends,  for  which  .-  had  the  value  154,  deflected  .01  in.  for  an 

d      A 
increase  in  the  load  of  3000  lbs.     Show  that      +  -^  =  .01  in. 

o         TT 

39.  A  long  column  with  pin  ends  is  bent  laterally  until  the  angular 
deviation  (5o)  at  the  ends  is  4°.  Find  the  total  maximum  intensity  of 
stress,  the  section  of  the  column  being  {a)  a  circle;  {h)  a  square. 
A"=  29,000,000  lbs.,  and  the  stress  due  to  direct  thrust  =  1500  lbs.  per 
square  inch.  Ans. — (a)  30,615  lbs.;  {b)  26,715  lbs. 

40.  With  the  same  vtaximuin  stress  as  in  the  last  question,  find 
the  angular  deviation  at  the  ends  so  that  the  stress  due  to  direct  thrust 
may  be  10,000  lbs.  per  square  inch.  Ans. — {a)  V  5';  {d)  i    33'. 

41.  Show  that  the  load  required  to  produce  an  angular  deviation  of 
14' at  the  two  pin  ends  of  a  long  column  is  on\y  one  per  cent  greater 
than  that  which  just  produces  flexure. 


m,  of  sectional 
and  that  the 


le  coefficient  of 


le  axis,  ^  being 


CHAPTER  IX. 


TORSION. 


I.  Torsion  is  the  force  with  which  a  thread,  wire,  or  pris- 
matic bar  tends  to  recover  its  original  state  after  having  been 
twisted,  and  is  produced  when  the  external  forces  which  act 
upon  the  bar  are  reducible  to  two  equal  and  opposite  couples 
(the  ends  of  the  bar  being  free),  or  to  a  single  couple  (one  end 
of  the  bar  being  fixed),  in  planes  perpendicular  to  the  axis  of 
the  bar.  The  effect  upon  the  bar  is  to  make  any  transverse 
section  turn  through  an  angle  in  its  own  plane,  and  to  cause 
originally  straight  fibres,  as  DE,  to  assume  helicoidal  forms,  as 
FG  or  DC.     This  induces  longitudinal  stresses  in  the  fibres. 


Fig.  360.  * 

and  transverse  sections  become  warped.  It  is  found  suf- 
ficiently accurate,  however,  in  the  case  of  cylindrical  and  regu- 
lar polygonal  prisms,  to  assume  that  a  transverse  section  which 
is  plane  before  twisting  remains  plane  while  being  twisted. 
In  order  that  the  bar  may  not  be  bent,  its  axis  must  coincide 
with  the  axis  of  the  twisting  couple. 

2.  Coulomb's  Laws. — The  angle  turned  through  by  one 
transverse  section  relatively  to  another  at  a  unit  distance  from 
it,  is  called  the  Angle  of  Torsion,  and  Coulomb  deduced  from 

568 


111! 


TORSIONAL    STRENGTH  OF  SHAFTS. 


569 


experiments  upon  wires,  that  this  angle  is  directly  proportional 
to  the  moment  of  the  twisting  couple,  and  inversely  propor> 
tional  to  the  fourth  power  of  the  diameter. 

Thus,  if  a  force  P,  at  the  end  of  a  lever  of  radius  /,  twists  a 
cylindrical  bar  of  length  L  and  radius  R^  and  if  Q  is  the  drcUf 
lar  measure  of  the  angle  of  torsion,  then 


Q  oc  Pp,  and  also  a 


R" 


so  that  6/  =  C-^,,  C  being  a  constant  depending  only  upon  the 

nature  of  the  material.  '  - 

Let  T  be  the  total  angle  of  torsion,  in  circular  measure, 
i.e.,  the  angle  turned  through  by  one  end  of  the  bar  relatively 
to  the  other.     Then 

L  R*' 

3.  Torsional  Strength  of  Shafts  (see  Art.  23,  Chap. 
IV). — Consider  a  portion  of  the  shaft  bounded 
by  the  planes  C£  and  MN,  Fig.  361.  It  is  kept 
in  equilibrium  by  the  couple  (P,  —P),  and  by 
the  elastic  resistance  at  the  section  AfN.  Hence, 
this  elastic  resistance  must  be  equivalent  to  a 
couple  equal  and  opposite  to  (P,  —P).  Fig.  36a. 

Let  Fig.  362  be  the  transverse  section  at  MN,  on  an  en- 
enlarged  scale,  and  let  abb' a'  be  any  elementary  area  (=  ^A^ 
{P,—P)  of  the  surface  bounded  by  the  radii  OA,  OB,  and  by 
the  concentric  arcs  aa',  bb'. 

Let  :r,  be  the  distance  of  ^A^  from  O. 

It  is  assumed,  and  is  approximately  true,  that  the  resist- 
ance of  any  element  abb' a'  to  torsion  is  directly  proportional 
to  the  angle  of  torsion  (^),  to  its  distance  from  the  axis  (;»:,), 
and  to  its  area  {^A,),  and  also  that  it  acts  at  right  angles  to 
the  radial  line  of  the  element,  i.e.,  to  OA  or  OB. 

Thus,  the  resistance  of  abb'a'  to  torsion  =  GBx^AA^ ,  G  be- 
ing a  constant  to  be  determined  by  experiment. 

The  corresponding  moment  of  resistance  about  the  axis  = 
GBx^AA^.      Similarly,    if    x^,  x^,  x^,   ...   are   the   distances 


f 


P.p 


%7o 


THEORY  OF  STRUCTURES. 


from  the  axis  of  any  other  elements,  J^,,   AA^,  AA^,... 
respectively,  the   corresponding   moments   of    resistance  are 
Gdx^AA^,  Gtix^AA^,  .  .  .  Hence,  the  Ma/  moment  of  resist- 
ance  of  the  section 

=  G(f{x,'JA,+x:dA,-{-...),        ;. 
=  G(^^{x'^A)  =  Gei,    V 

,  /being  the  moment  of  inertia  with  respect  to  the  axis. 
But  this  moment  of  resistance  {M)  is  equal  and  opposite  to 
the  moment  of  the  couple  (7^,  —  /')      Hence, 

'''-/;'/": :"^^.:----^:'-^     M=Gei=z]p.     ,'/,.'.  '-';'',, 

The  twisting  moment  will  of  course  vary  with  a  variable 
resistance,  and  the  last  equation  gives  its  mean  value. 

The  shaft,  however,  must  be  designed  (see  Cor.  4)  for  the 
maximum  couple  to  which  it  may  be  subjected,  and  the  moment 
of  this  couple  (=  J/,)  may  be  expressed  in  terms  of  the  mean 
by  the  equation  :^ 

« 
f4  being  a  coefficient  to  be  determined  in  each  case.    In  a  series 

of  experiments  with  different  engines,  Milton  found  that  /.i 
varied  from  1.3  to  2.1,  but  doubtless  the  variation  is  often  be- 
tween still  wider  limits. 

Cor.  1.  Let  /be  the  stress  at  the  point  farthest  from  the 
axis.     For  a  so/uil  round  shaft,  of  diameter  D, 

/  =  — ,     and    /=  G^-. 


.'.  M=Pp  =  GdI=  —^flT  =  .196/Z)'. 
Let  T"  be  the  total  torsion  in  degrees.    Then 


^~  L  180' 


TORSIONAL   STRENGTH  OF  SHAFTS.  57I 

and  hence  •■    •     ■        ■  i*  »  •'- 

■^~  Z   180   2' 

or 

Taking  the  following  mean  values  of  G  and/: 

Material.  G  ,  f 

Cast-iron 6,3CX),ooo  5,600 

Wrought-iron 10,500,000  7,200 

Steel 12,000,000  11,200 

„  =  9.87'°  for  cast  iron,  =  \2.jT°  for  wrought-iron,  =  9.3 T" 

(or  steel. 

Thus,  the  twist  is  1°  each  9.8  diameters  in  length  for  cast- 
iron,  each  12.7  diameters  in  length  for  wrought-iron,  and  each 
9.3  diameters  in  length  for  steel.  This  is  often  mucli  too  small, 
and  in  practice  the  twist  is  usually  limited  to  ■^^°  per  lineal  foot 
of  length.  For  a  hollow  round  shaft,  D  being  the  external  and 
Z),  the  internal  diameter,  , 

I^j^{D^-D:),    and    f=Ge^. 

•■•  ^  =  ^/  =  l6'^—D        =  -'^^-^  ~~D~^' 

If  the  thickness  {T)  of  the  hollow  shaft  is  small  compared 

with  D, 

D'-D;  =  IT  -{D-  2Ty  =  WT,  approximately, 

and  .; 

M=Pp=i.S7n'T. 


.1.;.' 


IM 


i  ] 

i  f'' 


r 


lU: 


r 


ill 


572 


THEORY  OF  STRUCTURES. 


The  use  of  compressed  steel  admits  of  shafts  being  made 
hollow.    For  a  solid  square  shaft,  H  being  the  side  of  thesquare, 


/  = 


6'' 


and  /,  the  stress  at  the  end  of  a  diagonal,  =  GQ  —- r 

^        H     6 


=  -^/^'  =  .2i6/H\ 


and 


GI~  GH' 


<= 


2/ 


4/2/^ 


In  these  results  it  is  assumed  that  Cr^l  =  -^  or  =     ,     . 

is  constant  at  different  points  of  the  cross-section,  which,  how- 
ever, is  only  true  for  circular  sections. 

In  non-circular  sections  the  stress  is  more  generally  greatest 
at  points  in  the  bounding  surface  which  are  nearest  to  the  axis 
and  least  at  those  points  which  are  farthest  from  the  axis. 
St.  Venant,  who  first  called  attention  to  this  fact,  gave  the  fol- 
lowing, amongst  others,  as  the   results  of  his  investigations. 

Designating  by  unity  the  torsional  rigidity  I  =  -^-1  of  a  shaft 

with  circular  section,  the  torsional  rigidity  of  a  shaft  of  equal 


sectional  area  is  .8863,  .8863  X 


yiF 


/;• 


ac- 


2n 

qr7»  7255.  or 

cording  as  the  section  is  a  square,  a  rectangle  with  sides  in 
the  ratio  «  to  i,  an  equilateral  triangle,  or  an  ellipse  whose 
major  and  minor  axes  are  2a  and  2b,  respectively. 

Cor.  2.  The  torsional  stress  per  unit  of  area  at  a  distance  x 
from  the  axis  is  G6x.  > 

Hence,  if  ^  =::  1  and  x  =^  1,  6"  is  the  force  that  will  twist  a 


.  > 


^m\ 


Hk 


TORSIONAL    STRENGTH  OF  SHAFTS. 


573 


being  made 
)f  thesquare, 


/2 


unit  of  area  at  a  unit  of  distance  from  the  axis  through  an  angle 
unity. 

Cauchy  found  analytically  that  in  an  isotropic  body  G  is 
two-fifths  of  the  coefficient  of  direct  elasticity. 

Experiments  indicate  that  G  is  about  three-eighths  or  one- 
third  of  the  coefficient  of  direct  elasticity. 

Cor,  3.  For  a  solid  cylinder,  Pp  = ,  R  being  the  radius, 

Pp 
and  therefore  R*  a  -^^.     If  the  shaft  is  to  have  a  certain  spea- 

Pp 
fed  stiffness,  i.e.,  if  #  is  fixed,  R^  a  y,-,  and  for  a  given  twisting 


^    or  =  -^y- j 
n,  which,  how- 

lerally  greatest 
rest  to  the  axis 
Irom  the  axis, 
|t,  gave  the  fol- 
investigations, 

l:^)  of  a  shaft 

1  shaft  of  equal 
IT> 


or 


ac- 


with  sides  in 
ellipse  whose 

It  a  distance  x 

It  will  twist  a 


moment  R*  a  ^.    Now  G  is  nearly  the  same  for  wrought-iron 

and  steel,  so  that  there  is  little  if  any  advantage  to  be  gained 
by  the  use  of  the  latter. 

After  passing  the  elastic  limit,  the  stress  varies  much  more 
slowly  than  as  the  distance  from  the  axis,  and  there  will  be  a 
partial  equalization  of  stress,  the  apparent  torsional  strength 
being  increased. 

Cor.  4,  In  any  transverse  ,>ection  of  a  solid  cylindrical  shaft, 
the  maximum  unit  stress 


/  = 


rft  D      16  M, 


1/,  being  the  moment  of  the  maximum  twisting  couple. 

This  relation  is  true  so  long  as  the  stress  does  not  exceed 
the  elastic  limit,  and  agrees  with  the  practical  rule  that  the 
diameter  of  a  cylindrical  shaft  subjected  to  torsional  forces  is 
proportional  to  the  cube  root  of  the  twisting  couple. 

The  rule  is  usually  expressed  in  the  form 


M,  =  KD\    so  that     K  = 


fJL 

i6- 


Wohler's  experiments  show  that  the  value  of  /  depends, 
to  some  extent,  upon  its  fluctuation  under  the  variable  twist- 


.:  'M 


574 


THEORY  OF  STRUCTURES. 


ing  moment.  Ordinarily  it  should  not  exceed  7200  lbs.  per  square 
inch  for  wrought-iron,  in  which  case  K  =■  ^\%^  X  ^  =  1414. 
{No/i\ — If  P,  is  the  torsional  breaking  weight, 


W" 


s 

SI 
I     I 


\ 

I 


is  the  coefficient  of  torsional  rupture.) 

Cor.  5.  Let  ^F'be  the  work  transmitted  to  a  shaft  of  D  in. 
diameter,  in  foot-pounds  per  minute,  N  being  the  correspond- 
insi  number  of  revolutions.     Then 

12  fF  =  inch-pounds  transmitted  =  2nMN  =  2tt  — 'iV 

/^ 

KD\, 

—  2n~ — N, 

M 
Since  M  =  mean  twisting  moment  =  — .     Hence, 

Let  HP  he  the  horse-power  transmitted  per  minute.    Then 
IV  =  33000  HP.     Also  for  u'roughi-iron  K  =  i|  g^  x  ^^^ 

TT  490  HP        _,,  ,  .. 

Hence  ix^ ^  =  D\     and  if     yw  =  1.43, 


V  ^v  ' 


If  J 


I 


a  formula  agreeing  with  the  best  practice  in  the  case  of 
wrought-iron  shafts  subjected  to  torsional  forces  only.  Such 
shafts  should,  therefore,  carry  no  pulleys. 

Cor.  6.  The  resilience  of  a  cylindrical  axle  is  the  product  of 
one  half  of  the  greatest  moment  of  torsion  into  the  correspond- 
ing angle  of  torsion.  ' 

Cor.  7.  It  often  happens  in  practice  that  a  shaft  (or  beam) 
is  subjected  to  a  bending  as  well  as  to  a  torsional  action. 


DISTANCE  liETlVF.EX    THE   BEARINGS  OF  SHAFTING.   575 

The  combined  bctidiiip;  and  twisting  moments  are  equiv- 
alent (Art.  8,  Chap.  IV)  to  the  moment 


M,  =  M,  +  ^M:  -t-  M':  =  J/,  («  +  )/n'  +  I). 

'vhere  Mi,  =  nM,,  M,,  being  the  bending  and  Mi  the  twisting 
moment  at  the  given  section. 

Hence,  remembering  that  the  maximum  twisting  moment 
J/,  is  equal  to  yiMt ,  we  have  for  a  wrought-iron  shaft, 

4goIfP 


/'('^  +  v^«'+0^^-  =  ^'. 


■  V  =  ,36  -\-  ,  this  becomes 

Z?  =  4i 


.  /HP 


a  formula  agreeing  with  the  best  practice  in  the  case  of  trans- 
mission with   bending,  as,  e.g.,  in  the  crank-shafts  of  marine 


engines. 


It  often  happens  that  ;/  has  a  still  larger  value,  as,  e.g.,  in 
the  case  of  head  shafts  properly  supported  against  springing. 
The  usual  formula  is  then 


I)  =  s 


V 


HP 

'N' 


corresponding  to  «  =  .72  -{- . 

4.  Distance  between  Bearings. — The  distance  between 
the  bearings  of  a  line  of  shafting  is  limited  by  the  considera- 
tion that  the  stiffness  of  the  shaft  must  be  such  as  will  enable 
it  to  resist  excessive  bending  under  its  own  weight  and  under 
any  other  loads  (e.g.,  pulleys,  wheels,  etc.)  applied  to  it. 
For  this  reason,  the  ratio  of  the  maximum  deviation  of  the 
axis  of  the  shaft  from  the  straight  to  the  corresponding  dis- 
tance between  bearings  should  not  exceed  a  certain  fraction 
whose  value  has  been  variously  estimated  by  different  writers. 

Let  /  be  the  distance  in  feet  between  bearings,  a  the 
diameter  of   the   shaft  in  inches,  w  the  weight  of   the  ma- 


ir 


:i 


576 


THEORY  OF  STRUCTURES. 


terial  of  the  shaft  per  cubic  foot,  and  let  the  applied  load  be 
equivalent  to  a  load  per  lineal  unit  of  length  in  times  that 
of  the  shaft.  Assume  a  stiffness  of  pgVo"'  ^"*^  ^^^'  ^^^  '^'^'s  of 
the  shaft  is  truly  in  line  at  the  bearings.  The  maximum  de- 
flection  of  the  shaft  is  given  by  the  formula  (Art.  3,  Ex.  8, 
Chap.  VII) 

_^_    I    {m  4-  i)(\veight  of  shaft)/' .  1728 
^  -  '385  EI 

1  ,     .nd*    I       ,  64    /• .  1728 

z=-^(m4-  l) ZVl — -jr =r » 

384^      '     •'    4    144       ftd*       £ 


D 
•*•  /  ""  100 


4    144 
I        (w  +  i)w  /* 


2E  d" 


or 


-i/ 


Ed' 


^ozj{tn  +0' 


ExAMPLii:. — For  wrought-iron,    E  =  3o,ooo,ocx)  lbs.  and 
w  =  480  lbs. 


»  /    d' 
.\l—  12.7a  / — —-. 


If  the  applied  load,  instead  of  being  uniformly  distributed 
is  concentrated  at  the  centre,  the  maximum  deflection 


=  Z>in. 


i_  {m  -\-  ^)( weight  of  shaft)/' .  1728 


192 


EI 


and  hence 


=/. 


Ed' 


ioozv{in  +  i)  ' 
Example. — For  wrought-iron  /=  8.5A  / — —^ . 


CYLINDRICAL   SPIRAL   SPRINGS. 


577 


5.  Efficiency   of  Shafting.— Let  it  require  the  whole  of 

the  driving  moment  to  overcome  the  friction  in  the  case  of  a 

shaft  of  diameter  d  and  length  L.     The  efficiency  of  a  shaft  of 

/ 
the  same  diameter  and  length  /  =  i  —  j- . 


But 


fnd 


—^  =  {Pf)  =  moment  of  friction 


wnd^    d 
/( L— 


lAwnd^ 


L, 


li)  being  the  specific  weight  of  the  material  of  the  shaft,  and  yu 
the  coefficient  of  friction.     Hence, 

I  Zif 


and  the  efficiency  =  i  —  2^-7-, 


6.  Cylindrical   Spiral  Spring. — Let  the  figure  represent 
a  cylindrical  spiral  spring  of  length  s,  supporting 
a  weight  W.      Consider  a  section  of  the  spring  at 
any  point  B. 

At  this  point  there  is  a  shear  W^and  a  torque 
Wy,  y  being  the  distance  of  B  from  the  axis  of 
the  spring,  i.e.,  the  radius  of  the  coil. 

The  effect  of  W  may  generally  be  neglected 
as  compared  with  the  effect  of  the  moment  Wy, 
and  it  may  be  therefore  assumed  that  the  spring 
is  under  torsion  at  every  point.  Let  there  be  n 
coils.    Then         ':,  . 


Hence, 


5=  2nyn,  approx. 


Fig.  363. 


r  being  the  radius  of  the  spring. 


578  THEORY  OF  STRUCTURES. 

The  elongation  of  the  spring  . 

_  2WfS      Syf      2nfn/ 

Wy  Wy'^S        Pnr'S 

The  work  done      = dS  =  '  ^     ,    = :p^ — . 

2  GTrr  4G 

A  weight  hung  at  the  lower  end  tends  to  turn  as  well  as 
lengthen  the  spring,  and  this  is  due  to  a  slight  bending  action. 

According  to  Hartnell, /=  60,000  lbs.  per  square  inch  for 
f-in.  steel,  /  =  50,000  lbs.  per  square  inch  for  ^-in.  steel,  and 
G  varies  from  13,000,000  lbs.  for  ^in.  steel  to  11,000,000  lbs. 
for  ^-in.  steel. 

Also  for  wire  less  than  f  in.  in  diameter, 


W  =  1^°°°^' 


y 


and  the  deflection  = 


Wny* 
288ooor* 


Example. — A  wrought-iron  shaft  in  a  rolling-mill  makes  50 
revolutions  per  minute  and  transmits  120  H.  P.,  which  is  sup- 
plied from  a  waterfall  by  means  of  a  turbine.  Determine  the 
diameter  of  the  shaft  (i)  if  the  maximum  stress  in  the  metal 
is  not  to  exceed  9000  lbs.  per  square  inch ;  (2)  if  the  angle  of 
torsion  is  not  to  exceed  ^°  per  lineal  foot. 

As  a  matter  of  fact,  the  diameter  of  the  shaft  is  3!  in.  at 
the  bearings  and  4  in.  in  the  intermediate  lengths.  What  are 
the  corresponding  maximum  inch-stresses  in  the  metal? 

Let  the  twisting  couple  be  represented  by  force  P  at  the 
end  of  an  arm  /.     Then 


.\Pp 


P  X  27rp  X  95  =  120  X  33.000  ft.-lbs 

120  X  33000 


First, 


2n  X  95 
126000  X  i^ 


126000 ,  ,.     126000    .  ,, 
ft.-lbs.  = X  I2m.-lbs. 


19 


19 


19 


and  hence 


P/.  _  Z""-^'  _  9000  X  22 

^P  -     16  -  16  X  7    ' 


D  =  3.56  in 


126000 
Second,  X  12  =  Pp  = 

'19  ^ 

n  I 


EXAMPLE. 


579 


32 


But  d  =  ~g  -  X  77  X  —  ;    take  G  =  10,500,000.    Then 
126000  X  12  _  10500000      22        I  I         I        2r? 

19  32  7        ISO         13         12  7 


Hence, 


Z>*=  689.45,    and    Z)=5.i2in. 


Third,  the  maximum  stresses  in  the  real  shaft  at  the  bear- 
ings  and  in  the  intermediate  lengths  are  respectively  given  by 


126000  stress      22        ,  „^. 

X  12  =  —^  X  y  X  (3l)', 


and 


19 
126000 


stress       22        ,  ^, 
,9      X.2  =  -^-XyX(4)-. 


From  the  former,  the  maximum  stress  =  7682  lbs.  per  sq.  inch. 
"       "    latter,      "  ♦'  "      =  6330   "      "     "      " 


t 


»■' 


■;v'-i"T'-" 


Kl  tail 


illi  i 


r  I 

s  ' 


580 


THEORY  OF  STRUCTUKES, 


EXAMPLES. 

1.  A  steel  shaft  4  in.  in  diameter  is  subjected  to  a  twisting  couple 
which  produces  a  circumferential  stress  of  15,000  lbs.  What  is  the  stress 
(shear)  at  a  point  i  in.  from  the  centre  of  the  shaft } 

Determine  the  twisting  couple.  Ans.  7500  lbs.;  23,571^  lbs. 

2.  A  weight  of  2j  tons  at  the  end  of  a  i-ft.  lever  twists  asunder  a 
steel  shaft  if  in.  in  diameter.  Find  the  breaking  weight  at  the  end  of  a 
2-ft.  lever,  and  also  the  modulus  of  rupture. 

Ans.   i^tons;  23,510  lbs. 

3.  A  couple  of  A'^ft.-tons  twists  asunder  a  shaft  of  diameter  d.  Find 
the  couple  which  will  twist  asunder  a  shaft  of  the  same  material  and 
diameter  id.  Ans.  %N. 

4.  Compare  the  couples  required  to  twist  two  shafts  of  the  same 
material  through  the  same  angle,  the  one  shaft  being  /  ft.  long  and 
d  in.  in  diameter,  the  other  2/  ft.  long  and  2d  in.  in  diameter. 

Compare  the  couples,  the  diameter  of  the  latter  shaft  being  -. 

Ans.  I  to  8;  32  to  i. 

5.  Ashaftisft.  long  and  4i  in.  in  diameter  is  twisted  through  an  angle 
of  2°  under  a  couple  of  2000  ft.-lbs.  Find  the  couple  which  will  twist 
a  shaft  of  the  same  material  20  ft.  long  and  T\  in.  in  diameter  through 
an  angle  of  2^°.  Ans.  12,288  ft.-lbs. 

6.  A  round  cast-iron  shaft  15  ft.  in  length  is  acted  upon  by  a  weight 
of  2000  lbs.  applied  at  the  circumference  of  a  wheel  on  the  shaft ;  the 
diameter  of  the  wheel  is  2  ft.  Find  the  diameter  of  the  shaft  so  that  the 
total  angle  of  torsion  may  not  exceed  2°.  Ans.  3.53  in. 

7.  A  wrought-iron  shaft  is  subjected  to  a  twisting  couple  of  12,000  ft.- 
lbs.  ;  the  length  of  the  shaft  between  the  sections  at  which  the  power  is 
received  and  given  off  is  30  ft. ;  the  total  admissible  twist  is  4°.  Find 
the  diameter  of  the  shaft,  fi  (page  570)  being  \,  and  m  10,000,000  lbs, 

Ans.  7.74  in. 

8.  A  wrought-iron  shaft  20  ft.  long  and  5  in.  in  diameter  is  twisted 
through  an  angle  of  2°.  Find  the  maximum  stress  in  the  material,  m 
being  10,500,000  ft.-lbs.  Ans.  3819.2  lbs.  per  sq.  in. 


i 


EXAMPLES. 


9.  A  crane  chain  exerts  a  pull  of  6000  lbs.  tangentially  to  the  drum 
upon  which  it  is  wrapped.  Find  the  diameter  of  a  wrought-iron  axle 
which  will  transmit  the  resultinp;  rmiplc,  the  effective  radius  of  the  drum 
being  7^  in. ;  the  safe  working  stress  per  square  inch  being  7200  lbs. 

Ans.  3.17  in. 

10.  Find  the  diameter  and  the  total  angle  of  torsion  of  a  12-ft. 
vvrought-iron  shaft  driven  by  a  water-wheel  of  20  H.  P.,  making  25 
revolutions  per  minute,  m  being  10,000,000  lbs.,  and  the  working  stress 
7200  lbs.  per  square  inch.  Ans.  5.6  in.;  2°. 2. 

11.  A  turbine  makes  114  revolutions  per  minute,  and  transmits  92 
H.  P.  through  the  medium  of  a  shaft  8  ft.  6  in.  in  length.  What  must  be 
the  diameter  of  the  shaft  so  that  the  total  angle  of  torsion  may  not  ex- 


ceed — ,  m  being  10,500,000  lbs.  ? 


Ans.  4.7  in. 


Determine  the  side  of  a  square  pine  shaft  that  might  be  substituted 
for  the  iron  shaft. 

12.  A  steel  shaft  20  ft.  in  length  and  3  in.  in  diameter  makes  200 
revolutions  per  minute  and  transmits  50  H.  P.  Through  what  angle  is 
the  shaft  twisted  ? 

A  wrought-iron  shaft  of  the  same  length  is  to  do  the  same  work  at 
the  same  speed.  Find  its  diameter  so  that  the  stress  at  the  circumference 
may  not  exceed  |  of  that  at  the  circumference  of  the  stee!  shaft. 

Ans.  2°. 6;  3.556  in. 

13.  A  vertical  cast-iron  axle  in  the  Saltaire  works  makes  92  revolu- 
tions per  minute  and  transmits  300  H.  P.;  its  diameter  is  10  in.  Find 
the  angle  of  torsion.  Ans,  .0144°  per  lineal  foot. 

14.  In  a  spinning-mill  a  cast-iron  shaft  8J  in.  in  diameter  makes  27 

1° 
revolutions  per  minute;  the  angle  of  torsion  is  not  to  exceed  — per 

lineal  foot.     Find  the  work  transmitted.  Ans.  62.19  H.  P. 

15.  A  square  wooden  shaft  8  ft.  in  length  is  acted  upon  by  a  force  of 
200  lbs.,  applied  at  the  circumference  of  an  8  ft. -wheel  on  the  shaft. 
Find  the  length  of  the  side  of  the  shaft,  so  that  the  total  torsion  may 
not  exceed  2°  {m  =  400000).  What  should  be  the  diameter  of  a  round 
shaft  of  equal  strength  and  of  the  same  material  ? 

Ans,  4.96  in.;  5.09in. 

16.  A  shaft  transmits  a  given  H.  P.  at  A'' revolutions  per  minute  with- 
out bending.     Find  the  weight  of  the  shaft  in  pounds  per  lineal  foot. 

/H.P.\! 


^1 

"1 

r 


582 


THEORY  OF  STRUCTURES. 


17.  The  working  stress  in  a  steel  shaft  subjected  to  a  twisting  couple 
of  1000  in.-tons  is  limited  to  11,200  lbs.  per  square  inch.  Find  its  diam- 
eter; also  find  the  diameter  of  the  steel  shaft  which  will  transmit  5000 
H.  P.  at  66  revolutions  per  minute,  yu  being  f.         Ans.  10  in.  ;  6.88  in. 

18.  A  wrought-iron  shaft  is  twisted  by  a  couple  of  10  ft.-tons.  Find 
its  diameter  {a)  if  the  torsion  is  not  to  exceed  1°  per  lineal  loo\.,{b)  if  the 
safe  working  stress  is  7200  lbs.  per  square  inch,     in  =  10,000,000  lbs. 

^«5.— (tf)  3.7  in. ;  (^)  5.7  in. 

19.  A  steel  shaft  2  in.  in  diameter  makes  100  revolutions  per  minute 
and  transmits  25  H.  P.  Find  the  maximum  working  stress  and  the  tor- 
sion per  lineal  foot,  jn  being  10,000,000  lbs.  Also  find  the  diameter  of  ;i 
shaft  of  the  same  material  which  will  transmit  100  H.  P.  with  the  same 
maximum  working  stress.  Ans.  io,o32^*'r ''^s. ;  .0478°;  3.17  in. 

20.  Tlie  crank  of  a  horizontal  engine  is  3  ft.  6  in.  and  the  connecting- 
rod  9  ft.  long.  At  half-stroke  the  pressure  in  the  connecting-rod  is  500 
lbs.     What  is  the  corresponding  twisting  moment  on  the  crank-shaft  ? 

Ans.  1716^  ft.-lbs. 

21.  If  the  horizontal  pressure  upon  the  piston  end  of  the  connecting 
rod  in  the  previous  question  is  constant,  find  the  maximum  twisting  mo- 
ment on  the  crank-shaft. 

^1  ■    A         sin  0  cos  6   \      .  ,    . 
Ans.  F[  sm  0  +  — _         —    ,   0  bemg  given  by 
\  ^/w"  —  sin'O/ 

n"  cos'O  +  w^sin"  G  cos''  0  —  i)  +  «'  sin*  6(1  +  sin"  6)  —  sin » 6  =  0 

where  n  =  J//"  =  V"- 

A^.B. — If  sin'O  is  neglected  as  compared  with  «', 

„  .    -  /       cos  G^ 
the  maximum  moment  =  /'smG(  i  -+- 


.    „/        cosG\ 
.mG(i+-^). 


G  being  very  nearly  72°. 

22.  Show  that  a  hollow  shaft  is  both  stiffer  and  stronger  than  a  solid 
shaft  of  the  same  weight  and  length. 

23.  Find  the  percentage  of  weiglit  saved  by  using  a  hollow  instead  of 
a  solid  shaft. 

Ans.  If  of  equal  stiffness  = 


200 
w'  -H  I 


If  of  equal  strength  =  100 


1" 


^/m\m^  -  1)  ) 


m  being  the  ratio  of  the  external  to  the  internal  diameter 
f  !  of  hoilow  shaft. 

24.  A  hollow  cast-iron  shaft  of  12  in.  external  diameter  is  twisted  by 
a  couple  of  27,000  ft.-lbs.  Find  the  proper  thickness  of  the  metal  so  that 
the  stress  may  not  exceed  5000  lbs.  per  square  inch.  Ans.  .619  in. 


EXAMPLES. 


5«3 


25.  The  external  diameter  of  a  hollow  shaft  is/ times  the  internal. 
Compare  its  torsional  strength  with  that  of  a  solid  shaft  of  the  same  ma- 
terial and  weight.  j^  ^^a  _  j 

■^"•f-  --z/-^ • 

p'  4-  I 

26.  If  the  solid  shaft  is  10  in.  in  diameter,  and  the  internal  diameter 
of  the  hollow  shaft  is  5  inches,  find  the  external  diameter  and  compare 
the  torsional  strengths.  '  ^;^j,  5  ^j  in. .  ^^  to  3. 

27.  A  hollow  steel  shaft  has  an  external  diameter  d  and  an  internal 

diameter  —  .     Compare  its  torsional  strength  with  that  of  {a)  a  solid 

steel  shaft  of  diameter  d\  {b)  a  solid  wrought-iron  shaft  of  diameter  d; 
the  safe  working  stresses  of  steel  and  iron  being  5  tons  and  3J  tons  re- 
spectively. Ans.—(a)-^^\{b)l\. 

28.  What  twisting  moment  can  be  transmitted  by  a  hollow  steel  shaft 
of  8  in.  internal  and  10  in.  external  diameter,  the  working  stress  being 
5  tons  per  square  inch  ?  Ans.  184*  in.-tons. 

29.  If/i  is  the  safe  torsional  working  stress  of  a  shaft,  and  /j  is  the 
safe  working  stress  when  the  shaft  acts  as  a  beam,  show  that  the  tor- 
sional resistance  of  the  shaft  is  to  its  bending  resistance  in  the  ratio  of 
2/,  to/,. 

30.  The  wrought-iron  screw  shaft  of  a  steamship  is  driven  by  a  pair 

of  cranks  set  at  right  angles  and  21.7  in.  in  length;  the  horizontal  pull 

upon  each  crank-pin  is  176,400  lbs.,  and  the  effective  length  of  the  shaft 

is  866  in.     Find  the  diameter  of  the  shaft  so  that  (i)  the  circumferential 

stress  may  not  exceed  9000  lbs.  per  square  inch  ;  (2)  the  angle  of  torsion 

1° 
may  not  exceed  — -  per  lineal  foot  ;  in  being  10,000,000  lbs.     The  actual 

diameter  of  the  shaft  is  14.9  in.     What  is  the  actual  \.ox%\ot\  ? 

Ans.—{\')  14.53  '"•;  (2)  14.89  in.;  (3)  total  torsion  =  5°.545- 

31.  The  ultimate  tensile  strength  of  the  iron  being  60,000  lbs.  per, 
square  inch,  find  the  actual  ultimate  strength  under  unlimited  repetitions 
of  stress.  Ans.  54,899  lbs.  (Unwin's  formula). 

32.  What  is  the  torsion  in  the  preceding  question  when  one  of  the 
cranks  passes  a  dead  point  ? 

33.  A  steel  shaft  300  feet  in  length  makes  200  revolutions  per  minute 
and  transmits  10  H,  P.  Determine  its  diameter  so  that  ihe  greatest 
stress  in  the  material  may  be  the  same  as  the  stress  at  tiie  circumference 
of  an  iron  shaft  i  in.  in  diameter  and  transmitting  500  ft. -lbs. 

Ans.  .807  in.  (=  \  in.) 

34.  Determine  the  coefficient  of  torsional  rupture  for  the  shaft  in 
Question  33,  10  being  the  factor  of  safety. 

35.  A  wrought-iron  shaft  in  a  rolling-mill  is  220  feet  in  length,  makes 
95  revolutions  per  minute,  and  transmits  120  H.  P.  to  the  rolls ;  the  main 
body  of  the  shaft  is  4  in.  in  diameter,  and  it  revolves  in  gudgeons  3f  in. 


584 


THEORY  OF  STRUCTURES. 


in  diameter.     Find  the  greatest  shear  stress  in  the  shaft  proper  anri  in 
the  portion  of  the  shaft  at  the  gudgeons.       Ans.  6330.2  lbs.;  7508  lbs. 

36.  Power  is  taken  from  a  shaft  by  means  of  a  pulley  24  inches  in 
diameter  which  is  i<eyed  on  to  the  shaft  at  a  point  dividing  the  distance 
between  two  consecutive  supports  into  segments  of  20  and  80  in. ;  the 
tangential  force  at  the  circumference  of  the  pulley  is  5500  lbs.  If  the 
shaft  is  of  cast-iron,  determine  its  diameter,  taking  into  account  the 
bending  action  to  which  it  is  subjected.  Ans.  4.7  in. 

37.  Show  that  the  resilience  of  a  twisted  shaft  is  proportional  to  its 


w.ight. 


Ans. 


Resilience  =  — 
m 


/'  Volume 


38.  If  a  round  bar  of  any  material  is  subjected  to  a  twisting  couple, 
show  that  its  maximum  resilience  is  two-thirds  the  maximum  resilience 
of  the  material. 

39.  Determine  the  diameter  of  a  wrought-iron  shaft  for  a  screw 
steamer,  and  the  torsion  per  lineal  foot;  the  indicated  H.  P,  =  1000,  the 
number  of  revolutions  per  minute  =  150,  the  length  of  the  shaft  frf)ni 
thrust  bearing  to  screw  =  75  ft.,  and  the  safe  working  stress  =  7200  lbs. 
per  square  inch.  Ans.  6.67  in.  ;  10. 5. 

40.  In  a  spinning-mill  a  cast-iron  shaft  84  ft.  long  makes  50  revolu- 
tions per  minute  and  transmits  270  H.  P.  Find  its  diameter  (i)  if  the 
stress  in  the  metal  is  not  to  exceed  5000  lbs.  per  square  inch ;  (2)  if  the 

1° 
angle  of  torsion  per  lineal  foot  is  not  to  exceed  — -. 

Also  (3)  in  the  first  case  find  the  total  torsion. 

Ans.  (I)  7.02  in. ;  (2)  10.23  m.  ;  (3)  28°. 8. 

41.  A  circular  shaft  is  twisted  beyond  the  limit  of  elasticity.  If  the 
equalization  of  stress  is  perfect,  show  that  for  a  given  maximum  stress  the 
twisting  couple  is  greater  than  it  would  be  if  the  elasticity  were  perfect, 
in  the  ratio  of  4  to  3. 

42.  Determine  {a)  the  profile  of  a  shau  of  length  /  which  at  every 
point  is  so  proportioned  as  to  be  just  abb  to  bear  the  power  it  has  to 
transmit  plus  the  power  required  to  Cf-r-ome  the  friction  beyond  the 
point  under  consideration.  Find  {b)  rtKi  efficiency  of  such  a  shaft,  and 
(c)  the  efficiency  of  a  shaft  made  up  of  a  series  of  n  divisions,  each  of 
uniform  diameter. 

Ans.  (a)  The  radius/  of  any  section  distant  x  from  the  driving  end 

X 

\sy=:re  3^  ,  r  being  the  radius  of  the  driving  end  and 
.  ,      L  the  length  of  a  shaft  of  uniform  diameter,  such  that 

the  whole  driving  moment  is  required  to  overcome  its 
,  .  own  friction. 


(l>)e^;{c) 


['  -  ^)"- 


EXAMPLES. 


585 


43.  A  steel  shaft  carries  a  5-ft,  pulley  midway  between  the  supports 
and  nialtes  6  revolutions  per  minute,  t'u  tangential  force  on  the  pulley 
being  500  lbs.  Taking  the  coefficient  ot  working  strength  at  11,200  lbs. 
per  square  inch,  find  the  diameter  of  the  shaft  and  the  proper  distance 
between  tlie  bearings. 

44.  A  steel  shaft  4  inches  in  diameter  and  weighing  490  lbs.  per  cubic 
(out  makes  100  revolutions  per  minute.  If  the  working  stress  in  the 
metal  is  11,200  lbs.  per  square  inch,  find  the  twisting  couple  and  the  dis- 
tance to  which  the  work  can  be  transmitted  ;  the  coefficient  of  friction 
being  .05,  and  the  efficiency  of  the  shaft  f . 

Am.   140,800  in. -lbs. ;  8228^  ft. 

45.  If  the  shaft  is  of  steel,  and  if  the  loss  due  to  friction  is  20  per  cent, 
lind  the  distance  to  which  work  may  be  transmitted,  /<  being  .05. 

Ans.  6582^  ft. 

46.  A  wrought-iron  shaft  220  ft.  between  bearings  and  4  in.  in  diam- 
eter can  safely  transmit  120  H.  P.  at  the  rate  of  95  revolutions  per 
minute.     What  is  the  efficiency  of  the  shaft  ?     (//  =  -^.~)       Ans.  .976. 

47.  The  efficiency  of  a  wrought-iron  shaft  is  ^  ;  the  working  stress  in 
the  metal  is  7200  lbs.  per  square  inch  ;  the  coefficient  of  friction  is. 125. 
How  far  can  the  work  be  transmitted  .'  Ans.  4320  ft. 

48.  A  spring  is  formed  of  steel  wire;  the  mean  diameter  of  the  coils 
IS  I  inch  ;  the  working  stress  of  the  wire  is  50,000  lbs.  per  square  inch  ; 
the  elongation  under  a  weight  of  iQjS^-  lbs.  is  2  inches;  the  coefficient  of 
transverse  elasticity  is  12,000,000  lbs.  Find  the  diameter  of  the  wire  and 
the  number  of  coils. 

49.  Find  the  weight  of  a  helical  spring  which  is  to  bear  a  safe  load  of  6 
tons  with  a  deflection  of  1  inch,  G  being  12,000,000  lbs.,  andy  60,000  lbs. 

50.  Find  the  time  of  oscillation  of  a  spring,  the  normal  displacement 

under  a  given  load  being  A.  /~Z 

Ans.  ny  —  ■ 

51.  Find  the  deflection  under  the  weight  W^of  a  conical  helical  spring 
iiiiof  circular  section  ;  (l>)  of  rectangular  section,  the  radii  of  the  e.xtreme 
coils  being  yi  andja ,  and  the  radial  distance  from  the  axis  to  a  point  of 
the  spring  at  an  angular  distance  <p  from  the  commencement  of  the  spiral 


Va  —  /? 
being  given  by  the  relation = 


0 

27r;; 


Ans.  (a) 


Gr* 


(H  =  number  of  coils.) 
(  =  -^^  .  'f->''  =  o  and  v>  =  H; 


6'  +  A'  IV 


-        S  and  A  being  the  sides  of  the  rectangular  section. 
52.  The  efficiency  of  an  axle  is  i ;  the  working  stress  in  the  shaft  is 
9000  lbs.  per  square  inch;  the  coefficient  of  friction  is  .lo.    How  far  may 
work  be  transmitted  ? 


1 

Iff 


I 


585a 


THEORY  OF  STRUCTURES. 


B 


Fig.  A  shows  the  distortion  produced  by  twisting  a  round  >i-in.  iron  bar. 
Fig.  B  shows  the  distortion  produced  by  twisting  a  square  J^-in.  iron  bar. 


DISTORTION  OF  IRON  BARS  BY  TWISTING.  iZ^b 


The  above  figures  show  the  distortion  produced  by  twisting  &\%yt  J^"  iron  bar. 


:,  .-it! 


^f  a' 


II 


5-^ 


CHAPTER   X. 
STRENGTH    OF  CYLINDRICAL  AND  SPHERICAL   BOILERS. 


Fig.  364. 


I.  Thin  Hollow  Cylinders ;  Boilers ;  Pipes. 

Let  r  be  the  radius  of  the  cylinder. 
Let  t  be  the  thickness  of  the  metal. 
Let  p  be  the  fluid  pressure  upon  each  unit 
of  surface. 

Let/  be  the  tensile  or  compressive  unit 
stress,  according  as  /  is  an  internal  or  exter- 
nal pressure.  ' 
'      Assume  (l)  that  the  metal  is  homogeneous  and  free  from 
initial  strain  ;           . 
"      (2)  that  t  is  small  as  compared  with  r ; 
f     (3)  that   the  pressures   are  uniformly   distributed  1 
'I                over  the  internal  and  external  surfaces; 
.V     (4)  that  the  ends  are  kept  perfectly  flat  and  rigid ; 
;}     (5)  that  the  stress  in  the  metal  is  uniformly  dis- 
tributed over  the  thickne.ss. 
The  last  assumption  is  equivalent  to  supposing  that  it  isj 
the   mean   circumferential    stress  which    is   governed   by  the] 
strength  of  the  metal,  while  in  reality  it  is  the  internal  or  maxi- 
mum circumferential  stress  which  is  so  governed. 

The  figure  represents  a  cross-section  of  the  cylinder  ofj 
thickness  unity. 

A  section  made  by  any  diametral  plane,  as  AB,  must  de- 
velop a  total  resistance  of  2tf,  and  this  must  be  equal  and! 
opposite  to  the  resultant  of  *^he  fluid  pressure  upon  each  half| 
i.e.,  to  ipr.     Hence, 


2tf  =  2pr,     or    //  =  pr. 


(0 


586 


THIN  HOLLOW  CYLINDERS;    BOILERS;    PIPES. 


587 


This  formula  may  be  employed  to  determine  the  btirsting, 

[  froof,  or  working  pressure  in  a  cylindrical  or  approximately 

cj'lindrical  boiler,  provided  that/",  instead  of  beinj^  the  tensile 

or  compressive  unit  stress,  is  some  suitable  coefficient  which 

has  been  determined  by  experiment.     If  rf  is  the  efificiency  of 

1  a  riveted  joint,  the  formula 

may  be  employed  to  determine  the  working  pressure  in  a  cylin- 
I  drical  or  approximately  cylindrical  boiler. 

In  ordinary  practice  the  values  of  //  and /"are  given  by  the 
I  (ollowing  table : 


Material. 

Joint, 

1 

/in  lbs.  per  sq.  in. 

ft'roughi-iron 

Sieel '. '.. 

Single-riveted 

Double-riveted 

Treble-riveted 

Single-riveted 

Double-riveted 

Treble-riveted 

•55 

•  7 

.8  to  .85 

•55 

•7 

.  8  to  .85 

8000  to  9000 

(4 
(  1 

12000  to  13000 

1  ( 

ii 

tt 

For  cast-iron  cylinders  the  working  value  of /may  be  taken 
latabout  2000  lbs.  per  square  inch. 

The  total  pressure  upon  each  of  the  flat  ends  of  the  cylinder 

-  -    .        .         =  7rr*p. 

[The  longitudinal  tension  in  a  thin  hollow  cylinder 

2nrt       2/'     ^  '' 

|iind  is  one  half  of  the  circumferential  stress/. 

Cor.  I.  Let  the  cylinder  be  subjected  to  an  external  pressure 
1^  as  well  as  to  an  internal  pressure  p.     Then 


/t=pr-p'r\ 


(3> 


r  being  the  radius  of  the  outside  surface  of  the  cylinder,    /is. 
htension  or  a  pressure  according  as  pr  %_p'r'. 


^1  I 
'   1 


:'^' 


588 


THEORY  OF  STRUCTURES. 


Generally,  r  —  r'  is  very  small,  and  the  relation  (3)  may  be 
written 

2.  Thick  Hollow  Cylinder.— If  t  is  large,  the  stress  is  no 
longer  uniformly  distributed  over  the  thickness.  Suppose  that 
the  assumptions  (i)  and  (3)  of  Art.  i  still  hold,  also  that  the 
cylinder  ends  are  free,  and  that  the  annulus  forming  the  section 
of  the  cylinder  is  composed  of  an  infinite  number  of  concentric 
rings.  Under  these  conditions  the  straining  of  the  cylinder 
cannot  affect  its  cylindrical  form.  Hence,  right  sections  of  the 
cylinder  in  the  unstrained  state  remain  planes  after  the  strain- 
ing, so  that  the  longitudinal  strain  at  every  point  must  be  the 
same.     Two  methods  will  be  discussed. 

First  Method. — Let  dx  be  the  thickness  of  one  of  the 
rings  of  radius  x,  and  let  dq  be  th«,  intensity  of  the  circum- 1 
ferential  stress. 

pr  —  p'r'  =  difference   between   the  total   pressures  from , 

within  and  without  =  total  circumferential  stress  =    /    dg. 

If  it  be  assumed  that  the  thickness  (=  r'  —  r)  remains  imj 
changed  under  the  pressure,  then  the  circumferential  extension  j 
of  each  of  the  concentric  rings  must  be  equal  to  the  same  con- 
stant quantity  \,  and  therefore 


dq  =  Edx 


2nx' 


E  being  the  coefficient  of  elasticity.     Hence, 

,  ,       E\   r'dx      E\^       t' 
pr  —p'r'  =  —    /     —  =  —  log,  -. 


r^      ^ 


Let  /  be  the  tensile  unit  stress.     Then /= -£ if  the 

elastic  limit  is  not  exceeded,  and  therefore 


pr-p'r'=^fr\og,-, 


ID 


V 


:| 


or 


THICK  HOLLOW  CYLINDER. 


589 


i'/i-n,allascomparedwi.h/;  and  hence/    ^^«^ 

fr       ■^2\-~Jr~)-     ■    .    (5) 
m  most  cases  which  oen^r  : 

&F  ^ero  in  equation  (5), 


^      A' ^-27)'     '     . 


P  •     •     •     •■    (6) 

cJ!!-rT'"''^  ^y-  ^^P^^ence  '^"'""^  °f  strength 

ano  J;  i^etlti?''   '"  ^''    ^^^^'^^    Mechanics,    obtains    by 

'f/ be  neglected.     Hence,         _ 

/  ^  2  7^ '  approximately, 
•'^■■"-aU  as  compared  with/ and  therefore 


k  :: 


590  THEOKY  OF  STh'UCTUKES. 

r  r     f\    ^  2/r 

an  equation  identical  with  (6). 

Second  Mf.thod. — Consider  a  ring  bounded  by  tlic  radii 
X,  X  -\-  dx,  at  any  point. 

Let  q  be  the  normal  (i.e.,  radial)  intensity  of  stress. 

Let /be  the  intensity  of  stress  tangential  to  the  ring. 
•'     s  "     "  "  "      "       perpendicular  to  the  plane 

of  the  ring. 

Let  tx,  /S,  y  be  the  corresponding  strains. 

Let  E  ?nd  ///  '  he  respectively  the  coefificients  of  direct  and 
lateral  elas Kxlj. 

Then,  since  E,/,  s  are  principal  stresses  (Chap.  IV), 

'^~  E        mE  '     ^  ~  ;£        iuE  '     ^  ~  E        mE  '  ^'^ 

But  Y  is  constant.  Also,  since  the  ends  are  free,  the  total 
pressure  on  a  transverse  section  is  nil,  and  hence  it  might  be 
inferred  that  s  is  zero  at  every  point.     Adopting  this  value  of  s, 

Byeq.  (I), 

...  f-\-q=.  a  constant  =c (2) 

Again, 

d{qx)  =  fdx  =  xdq -\- qdx (3) 

By  eqs.  (2)  and  (3), 

xdq -\- iqdx  =  cdx.  ' 

.•.  d{x*q)  =  cxdx. 
Integrating, 

^V  =  ~  +  ^,  .     .....    (4) 

(f  being  a  constant  of  integration. 

When  X  =■  r,  the  internal  radius,  q  '=^  p. 

"      X  =■  r\  the  external  radius,  q  z=  p\  _■'-.' 


Hence,  by  eq.  (4), 


SPHERICAL   SHELLS. 


59' 


and  therefore 


2 


/• 


^^^—' -^    and     c' ^^{tnjYr'* 


Hence,  by  eq.  (4), 


q  =  -S -^  -.P-P      r'r 


.»«/9 


and,  by  eq.  (2), 


r'  _  *' »     i — 


r   —  r 


,/  a  •       • 


•    (5) 


.    (6) 


3-  Spherical  Shells  —I  ^f  ^i      . 
T>.e  »«i„„  „„j,  ^  ^^^  "<=  "-  -me  as  before. 
«sta„cQ  of  2^«/.    Then  ^  "'"  """^  develop  a  total 


or 


W^pr.   . 


Hence,  a  spherical  «fi  n  •  '     *     *     vU 

longest  parts  of  ,a-.v./,.,/bol  ,  t° t       T'"''  ^^  "'^'  'he 

^^■^^  I.  Let  the  shf-M  k        !  ^"^"  ^'"'^■'^- 

''-  weU  a.  to  an  inte'™,";::^^-';'  ';,:„«'"-'  P--ure 

2     ^-f  ~  ^fp   ~  nr'y, 
,\f{r'  +  r)t  =,  r'p -^  r' y  -  '       ^^ ' 


/'  =  ?(;»-./.>  .   . 


(3) 


592 


THEORY  OF  STRUCTURES. 


Cor.  2.  For  a  thick  hollow  sphere,  Rankine  obtains 


P  =  2fzr^ 


r"  -  r' 


r"  +  2r' 


,  approximately. 


•    (4) 


4.  Practical  Remarks. — A  common  rule  requires  that  the 
working  pressure  in  fresh-water  boilers  should  not  exceed  one- 
sixth  of  the  bursting  pressure,  and  in  the  case  of  marine  boilers 
that  it  should  not  exceed  one-seventh. 

An  Ens^lish  Board  of  Trade  rule  is  that  the  tensile  workiii'^ 
stress  in  the  boiler-plate  is  not  to  exceed  6000  lbs.  per  square 
inch  of  gross  section,  and  French  law  fixes  this  limit  at  4250 
lbs.  per  square  inch. 

The  thickness  to  be  given  to  the  wrought-iron  plates  of  a 
cylindrical  boiler  is,  according  to  French  law, 

/  =  .oo2)6nr  -f-  .1  in.; 

according  to  Prussian  law,       ;; 

/  =  (^•°°3"  —  \)r  -\-  I  m.  =  .003«r  +  -i  in->  approximately, 

r  being  the  radius  in  inches,  and  n  the  excess  of  the  internal 
above  the  external  pressure  in  atmospheres. 

The  thickness  given  to  cast-iron  cylindrical  boiler-tubes  is, 
according  to  French  law,  five  times  the  thickness  of  equivalent 
wrought-iron  tubes ;  according  to  Prussian  law, 

f  _  ^^.oi«  _  jy  _|_  ^  in.  —  ,oinr  -\-  ^  in.,  approximately. 

Steam-boilers  before  being  used  should  be  subjected  to  a 
hydrostatic  test  varying  from  i^  to  3  times  the  pressure  at 
which  they  are  to  be  worked. 

Fairbairn  conducted  an  extensive  series  of  experiments 
upon  the  collapsing  strength  of  riveted  plate-iron  flues,  by 
enclosing  the  flues  in  larger  cylinders  and  subjecting  them  to 
hydraulic  pressure.  From  these  experiments  he  deduced  the 
following  formula  for  a  wrought-iron  cylindrical  flue  or  tube: 


Collapsing  pressure 
in  pounds  per  square  inch  of  surface 


I  =/ =403150-^- 


-^  =  403150 


a'  Ir' 


^y  riveting  an^Je-  or  t  •  .        ^ 

and  =:°°'^''*^  +  -°^-.«of.W..o„. 


^a/;-    or     £.^./ 


hccordfng  as  the  plate  is  m.    , 
'^^'■?'-d]y  fixed  around\r  ^^^^'"PP^'-^^d^'-ound  the  r'. 

I       -responding  deflections  ^th^platar^  "^  ^^^-^^^• 


61?;: 


and 


1(7)*^' 


ittl 


E' 


594 


THEORY  OF  HTKUCTUKES, 


EXAMPLES. 


1.  What  should  be  the  thickness  of  the  plates  of  a  cylindrical  boiler 
6  ft.  in  diameter  and  worked  to  a  pressure  of  50  lbs.  per  square  inch,  in 
order  that  the  working  tensile  stress  may  not  exceed  1.67  tons  per  square 
inch  of  gross  section  ?  Ans.  .42  in. 

2.  A  cylindrical  boiler  with  hemispherical  ends  is  4  ft.  in  diameter 
and  22  ft.  in  length.  Determine  the  thickness  of  the  plates  for  a  steam- 
pressure  of  4  atmospheres. 

3.  What  is  the  collapsing  pressure  of  a  flue  10  ft.  long,  36  in.  in 
diameter,  and  composed  of  i-in.  plates?  Also  of  a  flue  30  ft.  long,  48 in. 
in  diameter,  and  j\  in.  thick?  Ans.  490.84  lbs.;  91.59  lbs. 

4.  Determine  the  thickness  of  a  2-in.  locomotive  fire-tube  to  support 
pn  external  pressure  of  5  atmospheres. 

5.  A  copper  steam-pipe  is  4  in.  in  diameter  and  \  in.  thick.  Find  the 
working  pressure,  the  safe  coefficient  of  strength  for  copper  being  1000 
lbs.  per  square  inch.  Ans.  125  lbs.  per  square  inch. 

6.  A  7-ft.  boiler  of  -j'ls-in.  plates  was  burst  at  a  longitudinal  double- 
riveted  joint  by  a  pressure  of  310  lbs.  per  square  inch.  Find  the  coef- 
ficient of  ultimate  strength.  Ans.  29,760  lbs. 

7.  A  50-in.  cylindrical  boiler  of  -^  in.  plates  is  made  of  wrought- 
iron  wliose  safe  coefficient  of  strength  is  4000  lbs.  per  square  inch.  Find 
the  working  pressure.  Ans,  50  lbs.  per  squaie  inch. 

8.  A  lo-in.  cast-iron  water-pipe  is  subjected  to  a  pressure  of  250  lbs. 
per  square  inch.  Find  its  thickness,  the  coefficient  of  working  strength 
being  2000  lbs.  per  square  inch.  Ans.  \\  in. 

9.  A  steel  spherical  shell  36  in.  in  diameter  and  f  in.  thick  is  sub- 
jected to  an  internal  fluid  pressure  of  300  lbs.  per  square  inch.  Find  its 
coefficient  of  strength.  Ans.  7200  lbs. 

10.  A  thin,  hollow,  spherical,  elastic  envelope,  whose  internal 
radius  is  R,  was  subjected  to  a  fluid  pressure  which  caused  it  to  expand 
gradually  until  its  radius  became  R\ .     Determine  the  work  done. 

11.  The  plates  of  a  cylindrical  boiler  5  ft.  in  diameter  are  \  in.  thick. 
Find  to  what  pressure  the  boiler  may  be  worked  so  that  the  tensile  stress 
in  the  plates  may  not  exceed  \\  tons  per  square  inch  of  gross  section. 


''  M 


EXAMPLES. 


595 


12.  Show  that  the  assumption  of  a  uniform  distribution  of  stress  in 
the  thickness  of  a  cylindrical  or  spherical  boiler  is  only  admissible  when 
the  thickness  is  very  small. 

13.  A  metal  cylinder  of  internal  radius  r  and  external  radius  nr  is 
sjbjected  to  an  internal  pressure  of  p  tons  per  square  inch.  Show  that 
the  total   work   done   in   stretching  the   cylinder   circumferentially   is 

y-    ,  _  -  ft. -tons  per  square  foot  of  surface,  £  bemg  the  metal's  co- 

etficient  of  elasticity.  .... 

14.  The  cast-iron  cylinder  of  an  hydraulic  press  has  an  external 
diameter  twice  the  internal,  and  is  subjected  to  an  internal  pressure  of 
/tons  per  square  inch.  Find  the  principal  stresses  at  the  outer  and  inner 
circumferences.  Also,  if  the  pressure  is  3  tons  per  square  inch,  and  if  the 
internal  diameter  is  10  in.,  find  the  work  done  in  stretching  the  cylinder 
circumferentially,  £  being  8000  lbs. 

Am.  At  inner circumference,'V  =/.  a  thrust,  and/=  —  |/,a  tension. 
At  outer  circumference,  g  =  o,  and/  =  —  |/>,  a  tension. 

Work  =  126  ft.-lbs.  per  square  foot  of  surface. 

15.  The  chamber  of  a  27-ton  breech-loader  has  an  external  diameter 
of  40  in.  and  an  internal  diameter  of  14  in.  Under  a  powder  pressure  of 
18  tons  per  square  inch,  find  the  principal  stresses  at  the  outer  and  inner 
circumferences,  and  also  the  work  done  ;  £  being  13,000  lbs. 

Ans.  At  inner,  y  =  18  tons,  compression  ;  at  outer,  ^  =  o. 

At  inner,/ =  —  23^  tons,  tension  ;  at  outer,/ =  —  s^\  tons, 

tension. 
Work  =  li  ft.-tons  per  sq.  ft.  of  surface. 

16.  What  should  be  the  thickness  of  a  9-in.  cylinder  (a)  which  has 
to  withstand  a  pressure  of  800  lbs.  per  square  inch,  the  maximum  allow- 
able tensile  stress  being  24,000  lbs.  per  square  inch  ;  (d)  which  has  to 
withstand  a  pressure  of  6000  lbs.  per  square  inch  ;  the  maximum  allow- 
able tensile  stress  being  10,000  lbs.  per  square  inch  ? 

Afts. — (a)  1.86  in.  ;  (d)  4^  in. 

17.  Show  that  the  radial  (a)  and  hoop  (/3)  strains  in  thick  hollow 

cylinders  and  spheres  are  connected  by  the  relation  a  =  — ; — . 

ax 

18.  Prove  that  the  relation  in  Ex.  17  is  satisfied  by  the  values  ob- 
tained for/ and  g  in  the  Second  Method  of  Art.  2,  Chap.  X. 

19.  A  thick  hollow  sphere  of  internal  radius  r  and  external  radius 
'ir  is  subjected  to  an  internal  pressure  /  and  an  external  pressure  p . 
Determine  the  principal  stresses  at  a  distance  x  from  the  centre. 


Ans.  q  =■- 


«°—  I 


P  +  l_-I' 


«•—  I 


/ 


_p'n^-p     p-p'    n*r* 


Ml 


«•  -  I 


2X' 


«'  —  I 


I  if' 


m^': 


ill 


596 


THEORY  OF  STRUCTURES. 


20.  Assuming  tliat  the  annulus  forming  the  section  of  a  cylindrical 
boiler  is  composed  of  a  number  of  infinitely  thin  rings,  show  that  the 

pressure  at  the  circumference  of  a  ring  of  radius  r  is  —~  per  unit  of 


surface,  and  that  the  circumferential  stress  is  - 


B 


-,  A  and  B  de- 


r       mr'»  +  • 

noting  arbitrary  constants,  and  m  being  the  coefficient  of  lateral  con- 
traction. Find  the  values  of  A  and  B.po  and  pi  being  respectively  the 
internal  and  external  pressures. 

21.  Show  that  in  the  case  of  a  spherical  boiler  the  pressure  and  cir- 

cumferential  stress  are  respectively-^j^jj^  and  —  +  — :, .      Find 


fi  +  m 


(fn  -  Or"* -3 


A  and  B. 


22.  Solve  Questions  i,  2,  6,  7,  8,  9,  and  11  on  the  supposition  th?.c  /  is 
not  small  as  compared  with  r. 

23.  Taking/ =4000  lbs.  per  square  inch  and  £'  =  30,000,000  lbs,, 
Find  the  thickness  and  deflection  of  the  end  plates  of  the  boiler  in  Ques- 
tion 7.  .-''■■■'■■      ■■'•■?■  '  "■■■■ 


I-  Class 

eral  classes, 
tilcver  bridg 
(D)  Arched  I 
of  bridges  in 
I  2.  Compi 

Flanges  in  1 
sometimes  vJ 
claimed  that  J 
chord  a  slope, 
^ucii  a  truss 
flanges  and  or 
bolic  form  is  r 
nution  in  dept 
tion.     Again,  i 
tile  lower  part 
and  the  girder 
of  several  span- 
continuity  isne 
"lost  economic; 
uniform  throug 
^ange  at  any  p 
plates. 

3.  Depth  of 

ally  varies  from 
the  span.     It  is  1 


.  CHAPTER   XI. 

'  BRIDGES. 

I-  Classification Wr,- 1 

"lever  bridges  (A  ,.  ,„    fc)  I       '"'f'^""'^]  girders;  (H,  |^". 

ID)  Arched  bridges  (cti^-^f,  rxT"  ''"''^"  '^'•''  '  ^ ' ')^ 
»'  bridges  in  Classes  A  and  H  ol  "'"'■■'"  '^'"'P'"  "■<--at=i 

2-  Comparative  Advantages  nf  r 
Flanges  in  Girders  for  Brid^.  „f *^r"' /""  Horizontal 
»met,n,es  varied  for  the  sake  'f  '^'^  ^'^••-  ^P'l-  is 

*™ed  that  an  economy  otlteria,''^''";''""-  ""^  "  ''a'»o 
herd  a  slope,  as,  e.g.,  in'^.he  «se "f '  h!  %Tt  ''^  ^'^f^  "- 
^"cl,  a  truss  is  intermediate  beta'         ^"''"  ^"''S':  (Art.  ,„). 

W  c  form  ,s  not  well  suited  to  „1„  ^  '=''™''  "i-  Para- 

""•-n  in  depth  lessens  the  r  "sS  !  "-'™"ion,  and  a  din,  -• 
»»■     Again,  if  the  bottom  Ta' J       °'  ""^  ^'"^"  '"  Astor. 
*' lower  part  of  the  girder  is  ref,  ■!    !'"'"'■  ""  bracing  f„r 
■»d  the  girder  itself  mL.  b-    ndep'nT   ,"'''''"  ""™"  "™'-'° 
»' several  spans  any  advantage  wX,°    I  '°  """  '"  a  bridge 
»"fnuity  is  necessarily  lost     oli^"' ""^'"  ^'  ''"'^able  from 
;;;t  economical  form'^f  ^^e^     .S  ''''\''"^-  "'^  "est  and 
■"form  throughout,  and  fn  '  h U  '"  """''  "'^  "^'Pt"  is 

1  nge  at  any  point  f^  obtainld  h  '""''">-  ">ickness  of 

pia.es,  ""'^'"^d  by  .ncreasing  the  number  of 

f  Arie?f?om^«2^,:,::/"-'^  (Class  A).-The  depth  usu- 

*----sgenei;;r;:::--;(a-;v^^ 

give  large  girders 

597 


598 


THEORY  OF  STRUCTURES. 


an  increased  depth,  and  they  should,  therefore,  be  designed  to 
have  a  specified  strength.  If  the  span  is  more  than  twelve  times 
the  depth,  the  deflection  becomes  a  serious  consideration,  and 
the  girder  should  be  designed  to  have  a  specified  stiffness.  The 
depth  should  not  be  more  than  about  i^  times  the  width  of 
the  bridge,  and  is  therefore  limited  to  24  ft.  for  a  single  and  to 
40  ft.  for  a  double-track  bridge. 

4.  Position  of  Platform. — The  platform  may  be  supported 
either  at  the  top  or  bottom  flanges,  or  in  some  intermediate 
position.  In  favor  of  the  last  it  is  claimed  that  the  main  girders 
may  be  braced  together  below  the  platform  (Fig.  365),  while 
the  upper  portions  serve  as  parapets  or  guards,  and  also  that 
the  vibration  communicated  by  a  passing  train  is  diminished. 
The  position,  however,  is  not  conducive  to  rigidity,  and  a  large 
amount  of  metal  is  required  to  form  the  connections. 


Fig.  365. 


Fig,  366. 


The  method  of  supporting  the  platform  on  the  top  flanges 
(Fig.  366)  renders  the  whole  depth  of  the  girder  available  for 
bracing,  and  is  best  adapted  to  girders  of  shallow  depth. 
Heavy  cross-girders  may  be  entirely  dispensed  with  in  the  case 
of  a  single-track  bridge,  and  the  load  most  effectively  distrib- 
uted, by  hying  the  rails  directly  upon  the  flanges  and  vertically 
above  the  neutral  line.  Provision  may  be  made  for  side  spaces 
by  employing  sufficiently  long  cross-girders,  or  by  means  of 
short  cantilevers   fixed   to  the  flanges,  the  advantage  of  the 


POSITION  OF  PLATFORM. 


599 


former  arrangement  being  that  it  increases  the  resistance  to 
lateral  flexure,  and  gives  the  platform  more  elasticity. 

Figs.  367,  368,  369  show  the  cross-girders  attached  to  the 
bottom  flanges,  and  the  desirability  of  this  mode  of  support 
increases  with  the  depth  of  the  main  girders,  of  which  the  cen- 
tres of  gravity  should  be  as  low  as  possible.  If  the  cross-girders 
are  suspended  by  hangers  or  bolts  below  the  flanges  (Fig.  369), 
the  depth,  and  therefore  the  resistance  to  flexure,  is  increased. 


Fig.  36b. 


Fig.  369. 


In  order  to  stiffen  the  main  girders,  brakes  and  verticals, 
consisting  of  ang'e-  or  teeiron,  are  introduced  and  connected 
with  the  cross-girders  by  gusset  pieces,  etc. ;  also,  for  the  same 
purpose,  the  cross-girders  may  be  prolonged  on  each  side,  and 
the  end  joined  to  the  top  flanges  by  suitable  bars. 

When  the  depth  of  the  main  girders  is  more  than  about 
5  ft.,  the  top  flanges  should  be  braced  together.  But  the 
minimum  clear  headway  over  the  rails  is  i6  ft.,  so  that  some 
other  method  should  be  adopted  for  the  support  of  the  plat- 
form when  the  depth  of  the  main  girders  is  more  than  $  ft. 
and  less  than  16  ft. 

Assume  that  the  depth  of  the  platform  below  the  flanges  is 
3  ft.,  and  that  the  depth  of  the  transverse  bracing  at  the  top  is 
I  ft. ;  the  total  limiting  depths  are  7  ft.  and  19  ft.,  and  if  i  to  8 
is  taken  as  a  mean  ratio  of  the  depth  to  the  span,  the  corre- 
sponding limiting  spans  are  56  ft.  and  152  ft. 


6oo 


THEORY  OF  STRUCTURES. 


5.  Comparative  Advantages  of  Two,  Three,  and  Four 
Main  Girders.— A  bridge  is  {jreneraily  constructed  with  two 
main  girders,  but  if  it  is  crossed  by  a  double  track  a  tliird  is 
occasionally  added,  and  sometimes  each  track  is  carried  by  tw  0 
independent  girders. 

The  employment  of  four  independent  girders  possesses  the 
one  great  advantage  of  facilitating  the  maintenance  of  the 
bridge,  as  one-iialf  may  be  closed  for  repairs  without  int  pt- 
ing  the  traffic.  On  the  other  haiid,  the  rails  at  the  appr^  ichcs 
must  deviate  from  the  main  lines  in  order  to  enter  the  bridge, 
so  that  the  width  of  the  bridge  is  much  increased,  and  far 
more  material  is  required  in  its  construction. 

Few,  if  any,  reasons  can  be  ur^ed  in  favor  of  the  introduc- 
tion of  a  third  intermediate  girder,  since  it  presents  all  tin 
objectionable  features  of  the  last  system  without  any  corre- 
sponding recommendation. 

The  two-girder  system  is  to  be  preferred,  as  the  rails,  hy 
such  an  arrangement,  may  be  continued  over  the  bridge  with- 
out deviation  at  the  approaches,  and  a  large  amount  of  ma- 
terial is  economized,  even  taking  into  consideration  the  in- 
creased weight  of  long  cross-girders. 

6.  Bridge  Loads. — In  order  to  determine  the  stresses  in 
the  different  members  of  a  bridge  truss,  or  main  girder,  it  is 
necessary  to  ascertain  the  amount  and  character  of  the  load  to 
which  the  bridge  may  be  subjected.  The  load  is  partly  dead, 
partly  livt\  and  depends  upon  the  type  of  truss,  the  span,  the 
number  of  tracks,  and  a  variety  of  other  conditions. 

The  dead  load  increases  with  the  span,  and  embraces  the 
weight  of  the  main  girders  (or  trusses),  cross-girders,  platform. 
rails,  ballast,  and  accumulations  of  snow. 

As  to  the  live  load  see  Art.  19. 

7.  Trellis  or  Lattice  Girders. — The  ordinary  trellis  or 
lattice  girder  consists  of  a  pair  of  horizontal  chords  and  two 
series  of  diagonals  inclined  in  opposite  directions  (Fig.  370). 
The  system  of  trellis  is  said  to  be  single,  double,  or  treble,  ac- 
cording to  the  number  of  diagonals  met  by  the  same  vertical 
section. 


TRELLIS  OR  I^-rir^r^ 

UR  LATTICE  GIRDERS.  g^j 


p.^  f'o.  370. 

'.-I.  ^^"■'^^X!^t^'tx,^Z:^^^^^        ■'"'--  f-  the 
-^^'J  plates,  or  of  links  and  pir"'       ""••"  °'  "'"■=<'  ""d 


"^lir 


■^inmiir 


The  verticals  and  diagonal,  maJ'hr  ,  '"°-  "- 

°"!"  ^"""blc  section,  but  the  dht  ,  "'  ^"  ^  ^-  '■  H,  u,  or 
'  »">Kle  system  of  .reUis,  are  ,„  fv  "'  'k"''"^'  '"  ""^  "-  "' 
«'l.c  points  of  intersection  ^  ^''  ''"^'  "^'''^d  together 

,,,  A"  Objection  . o  this  da.  Of  girder  is  ,e  number  Of  .he 

•« t  .nr :hia"r;n';  t^,:";t  - ''"-ined  on  the  ass„.p. 
Attributed  be,v„een  the  diL,o„  T  "'"'"'  '«"™  ''^  equally 
"1'.ivalent  .0  the  subl.it  , bn  '^f'  ""  "^  "■-'  section,  Ihich 
'"'^tresses  in  the  several  bar"  ""■""  =^'™^  '"'  "■=  differ. 

<•  p  -  ;«  »  "e  the  permanent  load  concentrated  at  each  ape. 

'=t."'^^=7i«'-4-=tL  ''"="'■"«  '"-«  -t  the 

tnis  shearing  force  mu«f  h    ^ 
„,  """^""^  '"-""-tted  through  the  diag. 


onals. 


Hence,  the  stress  in  ab  due  to  th. 

aue  to  the  permanent  load 


4    ^^^  ^  =  g  zi'sec  ft 


602 


THEORY  OF  STRUCTURES. 


Again,  let  w'  be  the  live  load  concentrated  at  an  apex. 

The  greatest  shear  at  mn  due  to  the  live  load  occurs  when 
every  apex  between  a  and  7  is  loaded. 

This  shear  =  corresponding  reaction  at  i  =  \%w' ,  and  the 
stress  in  ab  due  to  the  live  load 

=  i  X  Ww'  sec  6*  =  Ww'  sec  B. 

Hence,  the  total  maximum  stress  in  ab  =  (|w  -f-  ||w')  sec  d. 

The  greatest  stress  of  a  kind  opposite  to  that  due  to  the 
dead  load  is  produced  in  ab  when  the  live  load  w'  is  concen- 
trated at  every  apex  between  I  and  b. 

The  shear  to  be  transmitted  is  then  2\w  due  to  the  dead  load, 
and  —  ^w'  due  to  the  live  load,  and  the  resultant  stress  in  ab 

=  ^{2^w  —  \^tv')  sec  6  =  {^zv  —  ^^zv')  sec  0. 

This  stress  may  be  negative,  and  must  be  provided  for  by 
introducing  a  counter-brace  or  by  proportioning  the  bar  to 
bear  both  the  greatest  tensile  and  the  greatest  compressive 
stress  to  which  it  may  be  subjected. 

The  stress  in  any  other  bar  may  be  obtained  as  above. 

The  chord  stresses  are  greatest  when  the  live  load  covers 
the  whole  of  the  girder,  and  may  be  obtained  by  the  method 
of  moments,  or  in  the  manner  described  in  the  succeedhig 
articles. 

In  the  above  it  is  assumed  that  the  members  of  the  girder 
are  riveted  together.  If  they  are  connected  by  pins,  each  of 
the  diagonal  systems  may  be  treated  as  being  independent. 

Thus,  the  system  i  2^^34567  transmits  to  the  supports 
the  stresses  due  to  loads  at  a,  3,  and  5. 

The  shear  due  to  the  dead  load,  transmitted  through  al), 

=  reaction  at  i  —  load  at  a  =  -w  —  w  =  —  . 

2  2  , 

w 
Hence,  *:he  stress  in  ab  due  to  the  dead  load  =  —  sec  6.  • 

2 

The  sircss  in  ab  due  to  the  live  load  is  greatest  when  tv'  is 

concentrated  at  each  of  the  points  3  and  5, 


The  maj 


and  the  corr 
Hence,  tl 


as  compared 
sumption. 
8.  Warre 

horizontal  ch. 
iinjjle  triangu 


The  princi 
girders  are  equ 

T'le  cross-gi 
the  apex  of  eac 

When  the  p 
sistance  of  the  ; 
'lorizontal  braci 
bracing  betweei 

VVlien    the  ] 

additional  cross 

tile  upper  chord 

rigidity  of  the  n 

Let  zu  be  the 

"  zu'  "    " 

"  /     "    " 

"  /k    "    " 

"     $      n      u 

"  ^'-f  I  be 


WARREN  GIRDER.  603 

The  maximum  shear  due  to  live  load  transmitted  through  ab 

=  if  w'  =  Jw', 

and  the  corresponding  stress  in  ab  =  ^w  sec  B, 
Hence,  the  total  maximum  stress  in  ab 


(t+H 


=  (~  +  ^«;']sece. 


as  compared  with  (Jzf/  + 1|^')  sec  B  obtained  on  the  first  as- 
sumption. 

8.  Warren  Girder. — The  Warren  girder  consists  of  two 
horizontal  chords  and  a  series  of  diagonal  braces  forming  a 
jinpfle  triangulation,  or  zigzag,  Fig.  375. 


n-l 


JS'+L 


N-1 


0  2  4  /t  N-2 

Fig.  375. 

The  principles  which  regulate  the  construction  of  trellis 
girders  are  equally  applicable  to  those  of  the  Warren  type. 

The  cross-girders  (floor-beams)  are  spaced  so  as  to  occur  at 
the  apex  of  each  triangle. 

When  the  platform  is  supported  at  the  top  chords,  the  re- 
sistance of  the  structure  to  lateral  flexure  may  be  increased  by 
horizontal  bracing  between  the  cross-girders  and  by  diagonal 
bracing  between  the  main  girders. 

When  the  platform  is  supported  on  the  bottom  chords, 
additional  cross-girders  may  be  suspended  from  the  apices  in 
the  upper  chords,  which  also  have  the  effect  of  adding  to  the 
rigidity  of  the  main  girders. 

Let  w  be  the  dead  load  concentrated  at  an  apex  or  joint. 

"    span  of  the  girder. 
"    depth  "    "        " 
"    length  of  each  diagonal  brace. 
"    inclination  of  each  diagonal  brace  to  the  ver- 
tical. 
iV-|-  1  be  the  number  of  joints. 


w 

I 

k 

s 

6 


€o4 


THEORY  OF  STRUCTURES. 


Then 


sec  5  =  ^,  tan  6*  =  ^. 


Two  cases  will  be  considered. 

Case  \.  All  the  joints  loaded. 

Chord  Stresses. — These  stresses  are  greatest  when  the  live 
load  covers  the  whole  of  the  girder. 

Let  5„  be  the  shearing  force  at  a  vertical  section  between 
the  joints  n  and  n-\-  i. 

Let  Hn  be  the  horizontal  chord  stress  between  the  joints 
«  —  I  and  n  •\-  \. 

The  total  load  due  to  both  dead  and  live  loads 

=  {w -\- w'){N  -  I). 
The  reaction  at  each  abutment  due  to  this  total  load 

w  -A-  w' 


The  shearing  forces  in  the  different  bays  are 

w  -\-  w' 
5„  =  {N  —  i),  between  o  and  i ; 


w  A-  v/ 
5.  =  — ~-(^-3). 


2; 


and 


5.  =  -^(^-5). 


'    w  A-  w' 
5.  =  —^-{N-  7), 


Sn=  -^-{N  -  2n  -  I). 


2  "    3; 

3  "   4; 


The  corresponding  diagonal  stresses  are 

Sa  sec  0,     S,  sec  0,     ....     S^  sec  6. 


WARREN  GIRDER. 


605 


The  last  stresses  multiplied  by  sin  B  give  the  increments  of 
the  chord  stresses  at  each  joint.     Thus, 

H^  =  tension  in  02  =  5„  tan  B 

w  +  w'  I    _w-\-w' I  N  —  \ 

H^  —  compression  in  i  3  =  5,  tan  8-j-  S,  tan  0 

_w-\-  w'  I  N  —  \       w-Vw'  I  N  —  3 
~        2       li~~N~^        2       li     N 

_w-\-w'  I  2{N—  2) 
~       2      ~k       N      ' 

H^  =  tension  in  2  4  =  //,  -|"  "S^i  tan  ^  +  6",  tan  0 

_w-\-  iv'  I  2>{N-i)  ,,  \ 

"  ~      2      k     n'    '  ■':"-■'':"■■■',::■■-'.:■ 

H^  —  compression  in  3  5  =  ^a  +  5,  tan  ^  +  5,  tan  B 

_zv  -\-w'  I  4{N  —  4)  _ 
~        2      ^       N      ' 

and  //„  =  horizontal    stress    in    chord,    between    the    joints 

w-\-  w'  /  n(N—  n)    ,    .  .        , 

n  —  I  and  n-\-  i  = y tz —  ,  being   a   tension   for   a 

'  2       k      N      '         ^ 

bay  in  the  bottom  chord,  and  a  compression  for  a  bay  in  the 

top  chord. 

Note. — The  same  results  may  be  obtained  by  the  method 
of  moments ;  e.g.,  find  the  chord  stress  between  the  joints 
"—  I  and  n  -{-  I. 

Let  a  vertical  plane  divide  the  girder  a  little  on  the  tight 
of  n. 

The  portion  of  the  girder  on  the  left  of  the  secant  plane  is 
kept  in  equilibrium  by  the  reaction  at  the  left  abutment,  the 
liorizontal  stresses  in  the  chords,  and  the  stress  in  the  diagonal 
from  «  to  «  -j-  I. 

Take  moments  about  the  joint  n.     Then 


i'i'il 


6o6 


THEORY  OF  STRUCTURES. 


2N 


w  4-  w' ,  N  —  n 
— ■ In- 


N 


.*.  H,t  =  etc. 


Diagonal  Stresses  due  to  Dead  Load. 

Let  d„  be  the  stress  in  the  diagonal  n,  n-\-  i,  due  to  the 
dead  load. 

The  shearing  forces  in  the  different  bays  due  to  the  dead 
load  are 


W  IV 

-{N  —  i),  between  o  and  i ;  -{N  —  3),  between  i  and  2; 

2  2 

W  W 

-{N-s\      "       2    "    4;  -(A^-7),       "        3    "   4; 


w 
and  —{N  —  2n  —  i),  between  n  and  «  -j-  i. 


The  corresponding  diagonal  stresses  are 


a  compression  -(N—  i)  sec  6  =.  -{N  —  iK  =  ^„  inoi 


2  ■ 


ZV  IV  s 

a  tension  -  (iV  —  3)  sec  9  =  —{N  —  3)t  =  df,  in  i  2 

w  w  s  . 

a  compression  —{N—  5)  sec  ^  =  —{N  —  5)r  =  a,  m  23 

2  2  ^ 


and  the  stress  in  the  »th  diagonal  between  n  and  n-\-iis 

s 


W  S 


"i 


WARREN  GIRDER. 


607 


being  a  tension  or  a  compression  according  as  the  brace  slopes 
down  or  up  towards  the  centre. 

Diagonal  Stresses  due  to  Live  Load. — The  live  load  produces 
the  greatest  stress  in  any  diagonal  («,  n-\-  \),  oi  the  same  kind 
as  that  due  to  the  dead  load,  when  it  covers  the  longer  of  the 
segments  into  which  the  diagonal  divides  the  girder.  Repre- 
sent this  maximum  stress  by  D„. 

The  live  load  produces  the  greatest  stress  in  any  diagonal 
(«,  n  -\-  l),  of  a  kind  opposite  to  that  due  to  the  dead  load,  when 
it  covers  the  shorter  of  the  segments  into  which  the  diagonal 
divides  the  girder.     Represent  this  maximum  stress  by  /?„'. 

The  shearing  force  at  any  section  due  to  the  live  load,  as  it 

crosses  the  girder,  is  the  reaction  at  the  end  of  the  unleaded 

segment,  and  the  corresponding  diagonal  stress  is  the  product 

s 
of  this  shearing  lorce  by  sec  B,  or  j- 

The  values  of  the  different  diagonal  stresses  are  :        ,  : 

/?,  =  compression  in  o  i  when  all  the  joints  are  loaded 

_  sw'N{N-  i) 
~  k  2        N      ' 


H-' 


m 


ZJ,  =  tension  in  1 2  when  all  the  joints  except  one  are  loaded 

_  sw'jN-  i){N-2) 

~  k  2  N  '  i 


Z),  =  compression  in  2  3  when  all  the  joints  except  i  and  2  are 

s  zv'  {N  -  2){N  -  3) 


loaded  = 


k  2 


N 


M 


D^  =  tension  in  34  when  all  the  joints  except  i,  2,  and  3  are 

k    2  N 

D^  =  stress  in  n,  n-\-  i  when  all  the  joints  except  i,  2,  3,  ,  .  . 

,      ,    ,       s  w' (N  —  n)(JV  —  n  —  1) 

and  n  are  loaded  =  r ^r ~ 

k  2  N 

A  =  stress  in  o  i  before  the  load  comes  upon  the  girder  =  o. 


ni 


6o8 


THEORY  OF  STKUVTUKES. 


S  W 


Z?/=  compression  in  i  2  when  the  joint  i  is  loaded     =  t^i. 


s  w 


DJ—  tension  in  2  3  when  the  joints  i  and  2  are  loaded  =  -r  \,x. 
'  k  N^ 

/?,'=  compression  in  34  when  the  joints  i,  2,  and  3  are  loaded 


/?„'=  stress  in  «,  «  +  I,  when  the  joints  1,2,...  and  ware  loaded 

s  7v'  n{n-\-  i) 
~  kN        2""' 

The  total  maximum  stress  in  the  «th  diagonal  of  the  same 
kind  as  that  due  to  the  dead  load  =  d^-\- D„. 

The  resultant  stress  in  the  «th  diagonal  when  the  load 
covers  the  shorter  segment  =:  d„—  D„'. 

This  resultant  stress  is  of  the  same  kind  as  that  due  to  the 
dead  load  so  long  as  d„  >  D„',  and  need  not  be  considered  since 
d„  -\-  D^  is  the  maximum  stress  of  that  kind. 

If  D^  >  d^,  it  is  necessary  to  provide  for  a  stress  in  the 
given  diagonal  of  a  kind  opposite  to  that  due  to  d„  +  -^-o  ^"<^ 
equal  in  amount  to  D^'  —  d„: 

This  is  effected  by  counterbracing  or  by  proportioning  the 
bar  to  bear  both  the  stresses  d„  -\-  D„  and  D„'  —  d„. 

Case  II.  Only  joints  denoted  by  even  numbers  loaded. 


0      2-4 


N-2    N 


Fig.  376. 


N-1 

N-2     N 


N~T 


Fig.  377. 

Chord  Stresses. — The  stresses  are  greatest  when  the  live  load 
covers  the  whole  of  the  girder. 


The  total  load  d 


"e  to  both  dead  and  J .Ve  loads 


609 


The 


'•eaetton  at  each  abutment 


zv  -\- 


To  find  H, ,  take 


'"e  to  this  totaUoad     .  - 


moments  about  i.     Th 


en 


wi.".: 


To  find /^r,  take 


AT  A  __  ^  +  ?f''  / 


moments  about  2. 


r  •  i. 


»«...  •  v--/- 


To  find  A^„  take 


^•^='-^'(^-.).^. 


moments  about  3. 


T°  find  ^,  take  „o„,ems  about  4.  . 


/  • 


4      ^^-2)4-^^^^.^^,^^_/^ 


,Jofind^.,  take  moments  aboutVand^.;,:t 


«  be 


even. 


4      ^^-2)«- 


"^"+"')^-K«~2)+.(«^ 


A^ 


and 


4)4-...+6  +  44.2f 
4  "~:j     -j, 


4    >^     iv^    ■• 


I 


6io 


THEORY  OF  UTRUCTURES, 


Next,  let  n  be  odd.     Then 


-(w  +  tt^)^{(»  -  a)-|-(«  -  4)  +  .  .  . -f  5-f  3+ ij 

/      .      '.  ^  ((i\^-2)«      {n  -  ly  ) 


and 


//-  = 


w-\-w'  I  (N—  2)n  -  («  -  I)' 
4      k  N 


Note. — If  —  is  even, 
2  ' 


w  4-  w'  / 
HiV,  the  stress  in  the  middle  bay,  =  — ^ —  tN 

N 
If  —  is  odd, 

"^  -L.  7^;'   /  ^'  ^ 

/T-v,  the  stress  in  the  middle  bay,  = v —  t — tt — . 

Diagonal  Stresses  due  to  the  Dead  Load. — The  shearing  forces 

w 
in  the  different  bays  due  to  the  d.ad  load  are  -{N  —  2)  between 

w  w 

O  and  2,  -(N  —  6)  between  2  and  4,  ~{N  —  10)  between  4  and 
4  4 

6,  etc. 

The  corresponding  diagonal  stresses  are 


S  IV , 


^,  in  o  I  =  T  -iN  —  2)    =  ^,  in  I  2 ; 


J  w 


^.  in  2  3  =  ^  7(^^-  4)    =  d,  in  34; 


jr  Tf*, 


d,mA^  =  j-^{N-  io)'^d,\n  56; 


etc.. 


etc., 


etc. 


The  truss  : 
combined. 

The  chord! 
"lore  parallel  r 
3"o\v  iron  susp 
^^'gs.  379  and 


HOWE   TRUSS. 

Thus  the  stresseq  in  f k    a-  ^** 

io'-;  are  e.ual  -Ca^  •  .::,t~  ^^^-H  n.eet  at  an  unloaded 

Diagonal  Stresses  ^ue  to  i^T r-^T^  '"  '^'"^- 
-  '"  Case  I.  and  ''  '''  ^^^^  ^''-^•^Thesc  are  found 


iV 


^^T'«<>dd.  there  is  a  single  stress  at'L     r 
these  coJumns.  '  ^^'  ^°°t  of  each  of 

The  maximum  rp<5iilfo.,4.    . 
'oads  /,  obuined  a/::?o':  '  ^'""  ''"=  ""  ""h  ^ead  and  „Ve 

^•g-»  the  maximum  resuJtanf  of 
segment  is  loaded  "^''"'  ^^''"^  '"  3  4  when  the  longer 

=  ^,  +  A  =  a;  +  /?„ 

and  the  maximum  resulfnnf   * 

-ent  is  loaded    "  "'"^^^"^  ^'^^  '"  34  when  the  shorter  seg- 

^-^  ^s  generally  6oo,  in  which  case  ;=  3/ 
9.  Howe  Truss.-Fiff   ,73  ,«        ,  ^ 

Howe  truss.  ^'  ^^«  '«  a  skeleton  diagram  of  a 


^''G-  378. 


'lie  chords     f  =  Kmi, 

7^  parallel  members  phc  J™' r"?"'"'y""^«'  of  three  .r 
^  '>on  s„spe„ders  w.Wew'ei''''!,''''''''""  -P"t  so"  "J 
(F'fis.  379  and  380).  '"'"'=''  '"d^  '<>  pass  between  them 


B^n 


612 


THEORY  OF  STRUCTURES. 


Each  member  is  made  up  of  a  number  of  lengths  scarfed 
or  fished  together  (Figs.  381  and  382). 

The  main  braces,  shown  by  the  full  diagonal  lines  in  Fig. 
378,  are  composed  of  two  or  more  members. 

The  counter-braces,  which  are  introduced  to  withstj^id  the 
effect  of  a  live  load,  and  are  shown  by  the  dotted  diagonal 
lines  in  Fig.  378,  are  cither  single  or  are  composed  of  two  or 
more  members.  They  are  set  between  the  main  braces,  and 
are  bolted  to  the  latter  at  the  points  of  intersection.  . 

The  main  braces  and  counters  abut  against  solid  hard-wood 
or  hollow  cast-iron  angle-blocks  (Fig.  380).  They  are  designed 
to  withstand  compressive  forces  only,  and  are  kept  in  place  by 
tightening  up  the  nuts  at  the  heads  of  the  suspenders. 


,-■ 

sA 

h'^ 

V 

■■ 

""' 

f- 

UV 

r^ 

^^ 

I 


-I — T 


-TT- 


^5^ 


Fig. 


379- 


Pig.  -do. 


Fig.  381. 


rr 


?-* V r^ 

Fig.  38a, 


hi  ZI 

jl  ■     Lil        ■ 


Fig.  383. 


Fit;.  384. 


The  angle-blocks  extend  over  the  whole  width  of  the  chords; 
if  they  are  made  of  iron,  they  may  be  strengthened  by  ribs. 

If  the  bottom  chord  is  of  iron,  it  may  be  constructed  on  the 
same  principles  as  those  employed  for  other  iron  girders.  It 
often  consists  of  a  number  of  links,  set  on  edge,  and  connected 
by  pins  (Figs.  383  and  3°  |.)-  I'l  such  a  case  the  lower  an;^le- 
blocks  should  hav^e  grooves  to  receive  the  bars,  so  as  to  prevent 
lateral  flexure. 

If  the  truss  is  made  entirely  of  iron,  the  top  chord  may  be 
formed  of  lengths  of  cast-iron  provided  with  suitable  flanges 
by  which  they  can  be  bolted  together.  Angle-blocks  may  also 
be  cast  in  the  same  p'oce  with  the  chord. 

To  determine   the  stresses  in   the  different  members,  the 


HOWE    TRUSS. 


613 


same  data  are  assumed  as  for  the  Warren  girder,  except  that 
A'' is  now  the  number  of  parcls. 

Chord  Stresses. — These  stresses  are  greatest  when  the  live 
load  covers  the  whole  of  the  girder. 

Let  //„  be  the  chord  stress  in  the  «th  panel. 

The  total  load  due  to  both  dead  and  live  loads 


=  {w-\-  zv'){N  —  i). 
The  reaction  at  each  abutment  due  to  this  total  load 

=  !!i±i":(Ar_,).     ■        ■ 


Let  a  plane  MM'  divide  the  truss  as  in  Fig.  378.  The  por- 
tion of  the  truss  on  the  loft  of  the  secant  plane  is  kept  in 
equilibrium  by  the  load  upon  that  portion,  the  reaction  at  the 
left  abutment,  the  chord  stresses  in  the  «th  panels,  and  the 
tension  in  the  «th  suspender. 

First,  let  the  load  be  on  the  top  chord  and  take  moments 
about  the  foot  of  the  «th  suspender.     Then 

iv-\-tv'  I  I  n{n  —  i) 

H„k  =  — 7— (A^  -  i)«^  -  {^v  +  zv')-^       3      ^ 


or 


H^ 


zi  -\-  zv'  n(N  —  n) 


w  -\-iv'  I  n{N  —  n) 
2~k        N       • 


NexJ,  let   the  Joad   be   on    the   bottom    chord    and  take 
moments  about  the  head  of  the  nth  suspender.     Then 


BJ 


zv  +  zc'    n{N  -  n) 

;; —  / T^: ,  as  before. 


Thus,  //■„  is  the  same   for  corresponding  panels,  whether 
the  load  is  on  the  top  or  bottom  cliord. 

Diagonal  Stresses  due  to  the  Dead  Load. — Let  VJ  be  the 


6i4 


THEORY  OF  STRUCTURES. 


shearing  force  in  the  «th  panel,  or  the  tension  on  the  «th  sus- 
pender due  to  the  dead  load. 

Fir&t,  let  the  load  be  on  the  top  chord.     Then 


F;'  =  --(A^-  \)-nw 


vi^-^  -  «). 


Next,  let  the  load  be  on  the  bottom  chord.     Then 
The  corresponding  diagonal  stresses  are 


and 


Diagonal  Stresses  due  to  the  Live  Load. — Let  V^'  be  the 
shearing  force  in  the  «th  panel,  or  tension  on  the  «th  suspen- 
der, when  the  live  load  covers  the  longer  segment. 

First,  let  the  load  be  on  the  top  chord. 

The  greatest  stress  in  the  «th  brace,  of  the  same  kind  as  that 
produced  by  the  dead  load,  occurs  when  all  the  panel  points 
on  the  right  of  MM'  are  loaded.  With  such  load,  Vn\  the 
shearing  force  on  the  left  of  MM',  =  the  reaction  at  o 

«''    ,  .N—  n 

* 

and  the  corresponding  diagonal  stress,  D„, 

N-n 


s 
k 


s  w 
k  2 


=  tV:'  ^T--^N-n-  I)- 


N 


Hence,  the  resultant  tension  on  the  «th  suspender  due  to 
both  dead  and  live  loads  =  F„  =  fV  +  V^' 


lN~x         \    ,   «;'    ^ 


N-n 

N    ' 


m 


HOWE   TRUSS. 


615 


and  the  resultant  maximum  compression  on  the  «th  brace  due 
to  both  dead  and  live  loads 

The  live  load  tends  to  produce  the  greatest  stress  in  the 
«th  counter  when  it  covers  the  shorter  segment  up  to  and  in- 
cluding the  «th  panel  point.  Even  then  there  will  be  no  stress 
in  the  counter  unless  the  effect  of  the  live  load  exceeds  that 
of  the  dead  load  in  the  («  -f"  0^^^  brace. 

The  shearing  force  on  the  right  of  MM'  =  the  reaction  at  N 


w'  n{n  -f-  i) 
~2  W 


Hence, 


s  w'  n(n  -f-  i) 
D„ ,  the  corresponding  diagonal  stress,  =  j ^—77 — -, 


*ii+i 


ind  the  resultant  stress  in  the  counter  =  D„'  — 
Next,  let  the  load  be  on  the  bottom  chord.     Then 


VJ'  =  -{N- 


n) 


N-  «  +  i 


and 


/?« 


s  w 
k   2 


{N-n) 


N 

N-n-\-\ 
N        ' 


Hence, 


K  =  K'  +  v."  =  .(^  -  „)  +  ^ V  _  „)^^=^. 

id 

^.  +  Z?,  =  ^1  w[-j-  -  «J  +  -{N  -  n) ^  J . 


Also,  the  stress  in  the  «th  counter  is 


d: 


s\iv'    n-\-i 


w 


^■-")[- 


6i6 


THEORY  OF  STRUCTURES. 


Note.— A  common  value  of  6  is  45°,  when  sec  6*  =  -7  =  1414, 

and  tan  6  =  -ry.  =  i . 

iVife  

The  end  panels  and  posts,  shown  by  the  dotted  lines  in 
Fig.  378,  may  be  omitted  when  the  platform  is  suspended  irom 
the  lower  chords. 

10.  Single  and  Double  Intersection  Trusses.  —  Fie,'. 
385    represents  the    simplest  form    of   single-intersection  (or 


l\ 

\.         / 

\^        y 

X 

\X 

/  \ 

/ J 

Fig.  385. 

Pratt)  truss ;  i.e.,  a  truss  in  which  a  diagonal  crosses  one  panel 
only.  It  may  be  constructed  entirely  of  iron  or  steel,  or  may 
have  the  chords  and  verticals  of  wood.  The  verticals  arc  in 
compression  and  the  diagonals  in  tension.  The  angle-blocks 
are  therefore  placed  above  the  top  and  below  the  bottom 
chord.  Counter-braces,  shown  by  the  dotted  diagonals,  are  in- 
troduced to  withstand  the  effect  of  a  live  load. 

If  the  truss  is  inverted  it  becomes  one  of  the  Howe  t\pe. 
and  the  stresses  in  the  several  members  of  both  trusses  may 
be  found  in  precisely  the  same  manner. 

Fig.  386  represents  a  double-intersection  (or  Whipple! 
^C2A'4     Cf    6    I     8 


H   7 
Fig.  386. 

truss,  i.e.,  a  truss  in  which  a  diagonal  crosses  fzvo  panels.  It 
may  be  constructed  entirely  of  iron  or  steel.  It  is  of  tlie  piii- 
connected  type,  and  the  two  diagonal  systems  may  be  treated 
independently. 

Let  B'  be  the  inclination  of  AB  to  the  vertical. 
"     e    "     "  "  "  A\,  CD,  .  ..  to  the  vertical. 

Chord  Stresses. — These  stresses  are  greatest  when  the  live 
load  covers  the  whole  of  the  girder. 

The  reaction  at  ^  from  the  system  ABCD .  .  .  =  4(w-|-ri'  1: 

"  "    /i         "  "  "  Al2i   .   .  .     =^{20^10  )\ 


SINGLE  AND  DOUBLE  INTERSECTION   TRUSSES.      617 

M'^xv'  being  the  dead  and  live  loads  concentrated  at  the  panel 
points  (7,  2,  £,  4, .  • . 

The  shearing  foces  in  the  different  bays  are :  •  ..^  ^  ^  •  ;' 

4(zy  -■}- w')  in  AC,  from  the  system  A  BCD  ;;",•'''-;  ^"  ■■ 


^123... 

ABCD... 

A12  ^  .  ,., 
ABCD ■ 

^123  ... 

ABCD 

A  I  2  ^  .  .  . 


^{w -\- w')  in  AC,    " 
3(w  +  w')  in  C2,      "        "        •' 
|(w  +  w')  in  2^,      "       '" 
2{w  +  w')  in  £4,      "        " 
|(w  +  w')  in  46^,      " 
i(w  +  w;')  in  (76,      "       " 
^{w  +  M/')  in  6/, 
The  corresponding  diagonal  stresses  are  : 
4{w  -\-  w')  sec  &'  in  AB ;  3i(zf  +  w')  ^ec  6  in  A  i  ', 
l{w  -{-  w')  sec  B  in  (7Z> ;     2|(z£'  +  ^'')  sec  0  in  23  ;     etc. 
Hence,  the  top  chord  stresses  are : 

C/m  AC  =  i\{w  -\-  zv')  tan  6'  -\-  il{zv  +  iv')  tan  6-, 

C,  in  C2  =  C,-\-  T){w  +  tc')  tan  6* 

=  4(w  -|-  2£;')  tan  6'  -\-  6^{w  -f-  w')  tan  6^ ; 

Cj  in  2E  =  C^-{-  2^{w  -\-  w')  tan  6 

=  4{w  -4-  w')  tan  6^'  -f-  9(w  -f-  ze;')  tan  6^ ;     etc. 

The  bottom  chord  stresses  are : 

r,  in  j5i  =  4(7€'  4-  w')  tan  ^' ; 

7;  in  iZ>  =  r,  +  i\(:cv  +  w')  tan  ^ 

=  df{w  -\-  w')  tan  0'  -\-  3i(w  +  w')  tan  6/  =  C, . 

So,        T,  =  C,,     Tt  =  C^,      etc.,  etc. 

Again,  the  stress  in  any  diagonal  4  5  of  the  system  A  i  2  . . . 
due  to  the  dead  load  =  i^w  sec  6*.  ,     .  •    ' 

The  live  load  produces  the  greatest  stress  in  4  5,  of  the  same 


\'t  >i 


i 


6i8 


THEORY  OF  STRUCTURES. 


kind  as  that  due  to  the  dead  load,  when  it  is  concentrated  at 
all  panel  points  of  the  system  ^  i  2  3  ...  on  the  right  of  4. 

The  reaction  at  A  is  then  ^w' ,  ^and  the  corresponding 
diagonal  stress  =  ^w'  sec  Q. 

Hence  the  maximum  resultant  stress  in  45  =  (|ze' -}--'/«/') 
sec  B. 

The  live  load  tends  to  produce  the  greatest  stress  in  any 
counter  5  8  when  it  is  concentrated  at  all  the  panel  points  of 
the  system  ^  i  2  3  ...  on  the  left  of  8. 

The  reaction  at  the  right  abutment  is  then  \w\  and  the 
corresponding  stress  in  the  counter  =  \w'  sec  B.  Thus,  the 
resultant  stress  in  the  counter  =  ^w' —\w)  sec  B,  \w  sec  Q  being 
the  stress  in  6  7  due  to  the  dead  load. 

Similarly,  the  stresses  in  any  other  diagonal  and  counter 
may  be  found. 

The  Pratt  truss  composed  entirely  of  iron  and  with  some 
of  the  details  of  the  Whipple  truss  is  sometimes  called  a 
Murpliy-W  hippie  truss.  The  Linville  truss  is  a  Whipple  truss 
made  of  wrought-iron,  the  verticals  being  tubular  columns. 

II.  Post  and  Quadrangular  Trusses. — The  peculiarity  of 

the  Post  truss  (Fig.  387)  is  that  the 
struts  are  inclined  at  an  angle  of 
about  22°  30'  to  the  vertical,  with  a 
view  to  an  economy  of  material. 
°  with  the  vertical. 


Fig.  388. 


Fig.  387. 

The  ties  cross  two  panels  at  an  angle  of  45 

In  the  quadrangular  truss 
(Fig.  388)  the  bottom  chord  has 
additional  points  of  support  half- 
way between  the  panel  points. 

The  Bollman,  Fink,  and  other  bridge-trusses  have  been 
referred  to  in  a  previous  chapter. 

12.  Bowstring  Girder  or  Truss. — The  bowstring  girder 
in  its  simplest  form  is  represented  by  Fig.  389,  and  is  an  excel- 
lent structure  in  point  of  strength  and  economy. 

The  top  chord  is  curved,  and  either  springs  from  shoes 
(sockets)  which  are  held  together  by  a  horizontal  tie,  or  has  its 
ends  riveted  to  those  of  the  tie.  t  :  -  /    ,  :■ 

The  strongest  bow  is  one  composed  of  iron  or  steel  cyHn- 


entrated  at 
?ht  of  4. 

rresponding 

^ress  in  any 
J  points  of 

v\  and  the 

Thus,  the 

sec  6  being 


BOWSTRmc   TRUSS.  '  ^ 

drical  tubes  hnf  o^         •    ,  ^^ 

•erhc-i  and  diagonals.  '"'"'"  '"  '^e  attachment  of 


"cal^blf  S  tr''''"fr''  '"""  "-=  bow  by  means  o, 

^  wnicn  are  usual  v  of  an  To    »•         ''^  '"eans  of  ver. 

ke  greatest  bread,!,  transvefse  so"  s  ,  "'  '""  ^--^  ^='  '^"h 

0  lateral  flexure.     In  large  br;d;es,,e""T-*  ""  ■■^^■»-" 

1'agonals  may  be  lattice-work  ""'  "'  ^"""1.  and 

If  the  load  upon  the  girri^r  ; 
;tat.o„ary,  verticals  only  are  renuLd"f°™'^  "'""'""«'  ^"d 
ke  neutral  axis  of  the  bow  should  be  t  ^"=^P"-°".  and 

"ly  distributed  load,  such  a  That  d,!   ,  ''^™''°'''-     ^n  irregu- 
»  change  the  shape  of  the  bow  and  d"  "  "T'"^  '^^'■"'  '"^^ 
to  resist  this  tendency.  '      ''  diagonals  are  introduced 

';,e-,bethedeadloadper,i„ealfoot.    "    ''^^ 

^  live      "       <<        „ 

.'.     1    '.'     ','     "P""  °f  the  girder. 

greatest  depth  ^^  of  .h,gr^^^ 

^'^ord  Stresses Thp 

ta  covers  the  whole  of",:  gS:;  "=  «-^'«'  -hen  the  live 


C^aWSi 


620  THEORY  OF  STRUCTURES. 

Let  Hht.  the  horizontal  thrust  at  the  crown. 
"     7*  "  .    •'  "  tension  in  the  tie. 

Imagine  the  girder  to  be  cut  by  a  vertical  plane  a  little  on 
the  right  of  BD.  The  portion  ABD  is  kept  in  equilibrium  by 
the  reaction  dX  A,  the  weight  upon  AD,  and  the  forces  H 
and  T.  '' 

Take  moments  about  B  and  D.     Then 


«;•::_ 


~  and 


"w  A-  iv' 


\  , 


'•    Let  H'  be  the  thrust  along  the  chord  at  any  point  P. 

Let  X  be  the  horizontal  distance  of  P  from  B. 

The  portion  PB  is  kept  in  equilibrium  by  the  thrust //at 
B,  the  thrust  H'  at  P,  and  the  weight  {w  -\-  w')x  between  P 
and  B.     Hence, 

.  ,     •  ^^.    .    ,    H'stc'i  =  H"  =  H'-{-{w-\-wYx\ 

i  being  the  inclination  of  the  tangent  at  P  to  the  horizontal, 

and  -  :  ,• 

the  thrust  at  ^  =  (-:^/)^^.+  ij   . 

..J  Diagonal  Stresses  due  to  Live  Load. — Assume  that  the  load  is 
concentrated  at  the  panel  points,  and  let  it  move  from  A 
towards  C. 


If  the  diagonals  slope  as  in  Fig.  390,  they  are  all  ties,  and 
the  live  load  produces  the  greatest  stress  in  any  one  of  them, 
as  QS,  when  all  the  panel  points  from  A  up  to  and  including 
Q  are  loaded. 


BOWSTRING   TRUSS. 


mt 


Let  X,  y  be  the  horizontal  and  vertical  co-ordinates,  respec-  . 
tively,  of   any  point  on  the  parabola  with  respect   to   B  as 
origin.  ,  u  .  ■.,,   _  .■  .  ■  . 

The  equation  of  the  parabola  is 


y  =  77^'. 


(I) 


Let  the  tangent  at  the  apex  P  meet  DB  produced  in  L, 
and  DC  produced  in  £. 

Draw  the  horizontal  line  PM.  -  •  --..-■  -    ■     ^ 

From  the  properties  of  the  parabola,  LM  =  2BM.        -        ' 
Let  PM  =  X  and  BM  =  y. 
From  the  similar  triangles  LMP  and  LDE, 


LM 

MP' 


LD 
DE' 


or 


21 

X 


k+y 


•    :'.QE=x" 


■y 


2y 


x+QE' 

P  -4x' 
Sx      ' 


Also 


,  c.  =  e.-(£-.)  =  <'-=i^. 


CE 
''QE 


I 


2X 


l-\-2x' 


(2) 


Draw  ^'F  perpendicular  to  QS  produced,  and  imagine  the 
girder  to  be  cut  by  a  vertical  plane  a  little  on  the  right  of  PQ. 

The  portion  of  the  girder  between  PQ  and  C  is  kept  in 
equilibrium  by  the  reaction  A'  at  C,  the  thrust  in  the  bow  at  P, 
the  tension  in  the  tie  at  (2,  and  the  stress  in  the  diagonal  QS. 

Denote  the  stress  by  D„ ,  and  let  the  panel  OQ  be  the  «th. 

Let  B  be  the  inclination  of  QS  to  the  horizontal.    _ 

Take  moments  about  E.     Then 


D^EF^R  X  CE, 


■:,ui.:.'   i  ;.. 


CE 
A  =  ^oE  ^^^^^  ^' ^^^ 


*  '■ 


622  THEORY  OF  STRUCTURES. 

^  ,     L.et  iV  be  the  total  number  of  panels.     Then   > 

I  I 

■TT^  is  a  panel  length,    and    tv'-jj  is  a  panel  weight. 

Also..  =  4- i,  and  hence 

•  .  r  a  • 

I  —  2x       CE  _N  —  n 


R,  the  reaction  at  C  when  the  n  panel  points  preceding  T 
are  loaded, 

_  w'  .n{n  -f-  i) 


\-\ 


Thus,  equation  (3)  becomes 


w'  N  —  n 

D„  =  — /(«  +  i)  — -r-^—  cosec  6. 


N' 


(4) 


Again,  by  equation  (i). 


.'.ST=ki 


I  -4 


n-\- 1        I 


-i)}  =  4^' 


(N-n-  !)(«+  I) 


A^= 


and 


cosec  e=-^  = 5jr— . 


Hence,  finally, 


'/ivr-.^+^u^-«-o(«+i)rJ_  ^^^ 


8  /&      N 


N—  n—  1 


This  formula  evidently  applies  to  all  the  diagonals  between  D 
and  C. 


V-  } 


BOWSTRING   TRUSS. 


623 


Similarly,  it  may  be  easily  shown  that  the  stress  in  any 
diagonal  between  D  and  A  is  given  by  an  expression  of  pre- 
cisely the  same  form. 

Hence,  the  value  of  D^  in  equation  (5)  is  general  for  the 
whole  girder. 

A  load  moving  from  C  towards  A  requires  diagonals  in- 
clined in  an  opposite  direction  to  those  shown  in  Fig.  390. 

Stresses  m  the  Verticals  due  to  the  Live  Load. — Let  F„  be  the 
stress  in  the  «th  vertical  PQ  due  to  the  live  load.  This  stress 
is  evidently  a  compression,  and  is  a  maximum  when  all  the 
panel  points  from  A  up  to  and  including  0  ar^  loaded. 

Imagine  the  girder  to  be  cut  by  a  plane  S' S"  very  near  PO, 
Fig.  390.  The  portion  of  the  girder  between  S S"  and  C  is  kept 
in  equilibrium  by  the  reaction  R'  at  C,  the  thrust  in  the  bow 
at  P,  the  tension  in  the  tie  at  O,  and  the  compression  V^  in 
the  vertical. 

Take  moments  about  E.    Then 


V^QE  =  R'X  CE,    or     F„  =  R' 


N^ 


n 


n 


and  R' ,  the  reaction  at  C  when  the  («  —  i)  panel  points  from  A 
to  and  ii 
Hence, 


w'  tiin i^ 

up  to  and  including  0  are  loaded,  =  — /■      „  -- 


(6) 


a  general  formula  for  all  the  verticals. 

Let  v„  be  the  tension  in  the  «th  vertical  due  to  the  dead 
load.  The  resultant  stress  in  it  when  the  live  load  covers  AO 
isz'„  —  V„,  and  if  negative,  this  is  the  maximum  compression 
to  which  PQ  is  subjected. 

If  v„  —  V„  is  positive,  the  vertical  PQ  is  never  in  compression. 

The  maximum  tension  in  a  vertical  occurs  when  the  live 

/ 

load  covers  the  whole  of  the  girder  and  =  7£''      -f-  the  tension 

due  to  the  dead  load. 

iVote. — The  same  results  are  obtained  when  JV  is  odd. 


ii)l 


I.  f  mn 


624 


THEORY  OF  S7KUCTUKES. 


13.  Bowstring  Girder  with  Isosceles  Bracing. 

I)itii;;<)nat  StrtsSiS liiii'  to  the  Dead  Load. — Uiuicr  a  iload  i.ui 
the  bow  is  equilibrated  aiul  the  tic  is  subjected  to  a  unidim 
tensile  stress  equal  in  amount  to  the  horizontal  thrust  at  the 
crown.  The  braces  merely  serve  to  transmit  the  load  to  ilu: 
bow  and  are  all  ties. 

Lot  Z, ,  T',  be  the  tensile  stresses  in  the  two  diagonals 
meeting  at  any  panel  point  Q.  Let  <^, ,  ^^  be  the  inclinations 
of  the  diagonals  to  the  horizontal. 

Let  W  be  the  panel  wei^jht  suspended  from  Q. 


--->*.£ 


Fig.  391. 


The  stress  in  the  tie  on  each  side  of  Q  is  the  same,  and 
therefore  T^,  T^,  and  ^Fare  necessarily  in  equilibrium. 
Hence,  , 

T  -  IV       ^°^  ^'  A      T  -  W—^^'— 

^'~  "^^sinl^.  +  ^J'     ''"''     ^'-  ''^sin{H,-\-Hy 
Dia£■OHa/  Stresses  due  to  the  Live  Load. — Let  iV^be  the  num- 
ber of /;«//"  panels. 

2/ 
The  length  of  a  panel  =  -rj  ;  the  weight  at  a  panel  point 

'  .2/' ^■-.7.:  '>"-7 : 


=  w 


N 


Let  the  load  move  from  A  towards  C.    All  the  braces  in- 
clined like  OP  are  ties,  and  all  those  inclined  like  QP  are  struts. 

The  live  load  produces  the  greatest  stress  in  OP  when  it 
covers  the  girder  between  A  and  O. 

Denote  this  stress  by  D„ ;  OG  is  the  «th  half-panel. 

As  before, 

D„EF=R  X  CE.   .    .     .     .    .    .    .    (i) 


ifc'tt 


BoiyarniNG  girder  with  isosceles  BRAChva.    625 

The  load  upon  AO  =  nxv'-rt,  and  hence  R  =  -j^ 2]V~ ' 

The  ratio  of  CJi  to  EF  is  denoted  by  the  same  expression 
as  in  the  preceding  article.     Thus,  , 


IK  = 


xv'  I  n-\-2N 


8  /fe  «  -I-  1     N 


"  ^  — ^.  (2) 


N  —  n  —  I 


The  live  load  produces  the  greatest  stress  in  OAf  when  it 
covers  the  girder  up  to  and  including  /?. 

Denote  the  stress  by  D„'  ;  BG  is  now  the  «th  half  panel. 

Let  K'  be  the  reaction  at  C. 

As  before,  -  .    . 


CE 
DJ  =  R'-^cosccO, 

I)  being  the  angle  MOD. 

I 
The  weight  upon  AD  =  (« —  i)w'^, 

and  hence  "• 


(3) 


R' 


w'  n*  —  I 
2    "W^- 


III 


It  may  be  easily  shown,  as  in  the  preceding  article,  that 


CE  _N_ 
0E~ 


n 


n-\-  I 


,  and  cosec  0  =  N 


[/■+J^i(^-»)»r]* 


4«^(iV  —  n) 


.-./?'  =  — 


w„-,iv-«-,[''+^V-«>rJ 


%  k      n 


N 


N-n 


(4) 


Hence,  when  the  load  moves  from  A  towards  C,  eq.  (2) 
gives  the  diagonal  stress  when  n  is  even,  and  eq.  (4)  gives  the 
stress  when  n  is  odd. 

If  the  load  moves  from  C  towards  A,  the  stresses  are  re- 
versed in  kind,  so  that  the  braces  have  to  be  designed  to  act 
both  as  struts  and  ties.  ,,,  ,     - 


626 


THEORY  OF  STRUCTURES. 


Note. — By  inverting  Fig.  391,  a  bowstring  girder  is  obtained 
with  the  horizontal  chord  in  compression  and  the  bow  in 
tension. 

14.  Bowstring  Suspension  Bridp;e  {Lenticular  Truss). — 
This  bridge  is  a  combina;;ion  of  tlie  ordinary  and  inverted 
bowstrings.  The  most  important  example  is  that  erected  at 
Saltash,  Cornwall,  which  has  a  clear  span  of  445  feet.  The 
bow  is  a  wrought-iron  tube  of  an  elliptical  section  stiffened 
at  intervals  by  diaphragms,  and  the  tie  is  a  pair  of  chains. 

A  girder  of  this  class  may  be  made  to  resist  the  action  of  a 
passing  load  either  by  ':he  stiffness  of  the  bow  or  by  diagonal 
bracing. 


In  Fig.  392,  let  BD  =  k,  B'D  =  k' , 

Let  //"be  the  horizontal  thrust  at  B,  and  T  the  horizontal  pull 
at  B' ,  when  the  live  load  covers  the  whole  of  the  girder.     Then 

_  w^iv'    r    _ 

First,  let  k  —  k' .     Then 

w  +  w'  /' 


//'=  T  = 


16      k' 


which  is  one  half  of  the  corresponding  stress  in  a  bowstrini; 
girder  of  span  /  and  depth  /•. 

(>.ie  half  of  the  total  load  is  supported  by  the  bow  and  ono 
half  is  transmitted  through  the  verticals  to  the  tie.     Hence, 


the  stress  in  each  vertical  —  -j;j{^v'  -|-  w"), 

zu"  being  the  portion  of  the  dead  weight  per  lineal  foot  borne 
by  the  verticals,  and  iV"the  number  of  panels. 


CANTILEVER    TRUSSES. 


627 


The  diagonals  are  strained  only  under  a  passing  load. 

Let  PP'  be  a  vertical  through  E,  the  point  of  intersection 
of  any  two  diagonals  in  the  same  panel,  and  let  the  load  move 
from  A  towards  0. 

By  drawing  the  tangent  at  P  and  proceeding  as  in  Art.  13, 
the  expression  for  the  diagonal  stress  in  QS  becomes,  as  before. 


%v'  n{n-  \)l-2x 
Similarly,  the  stress  in  the  vertical  QQ'  is 


(I) 


N 


N'        l-^2X 


(2) 


Next,  let  k  and  k'  be  unequal. 

Let  ^Fbe  the  weight  of  the  bow,  W  the  weight  of  the  tie. 

Then,  under  these  loads, 


'         _    TT  TTf    _     _ 

8    k~  ~"8"y6' 


w 


k 
~k" 


.     (3) 


The  verticals  are  not  strained  unless  the  platform  is  attached 
to  them  along  the  common  chord  ADO.  In  such  a  case,  the 
weight  of  the  platform  is  to  be  included  in  W' 

The  tangents  at  /'and  P'  evidently  meet  AO  produced  in 
the  same  point  O' ,  for  EO'  is  independent  of  k  or  k' .  Hence, 
the  stresses  in  the  verticals  and  diagonals  due  to  the  passing 
load  may  be  obtained  as  before. 

15.  Cantilever  Trusses. — A  cantilever  is  a  structure  sup- 
ported at  one  end  only,  and  a  bridge  of  which  such  a  structure 
forms  part  may  be  called  a  cantilever  bridge.     Two  cantilevers 


33- 


BRIDQEOVER  ST.  LAWRENCE  AT  NIAGARA. 
Fig.  393. 

may  project  from  the  supports  so  as  to  meet,  or  a  gap  may  be 
left  between  them  which  may  be  bridged  by  an  independent 


628 


THEORY   OF  STRUCTURES. 


girder  resting  upon  or  hinged  to  the  ends  of  the  cantilevers. 
The  form  of  the  cantilever  is  subject  to  considerable  variation. 


^^ 


SUKKL'R  BRIDGE 
Fig.   394, 


FORTH  BRIDGE. 
Fig.  395. 

Fig.s.  396  to  401  represent  the  simplest  forms  of  a  cantilever 
frame.     If   the  membei  AB  has  to  support  a  uniformly  dis- 


Fig.  396. 


Fig.  397. 


Fig.  398. 


N\KI\M\ 

A  B 

Fig.  400. 


Pig.   401, 


Fig,  402. 


. /f^TTVfv?^. 


Fig.  403.  Fio.  404. 

tributed  load  as  well  as  a  concentrated  load  at  />',  intermediate 
stays  may  be  introduced  as  shown  by  the  full  or  by  the  dt)ttod 


CA  N  TILE  VKR    Tk'  USSES. 


629 


linos  in  Figs.  398  and  399.  Should  a  live  load  travel  over 
/i;V,  each  stay  must  be  designed  to  bear  with  safety  the 
maximum  stress  to  which  it  may  be  subjected. 

Figs.  400  and  401  show  cantilever  trusses  with  parallel 
chords.  If  the  truss  is  of  the  double-intersection  type,  Fig. 
401,  the  stresses  in  the  members  terminating  in  Ji  become  in- 
determinate. They  may  be  made  deternn'nate  by  introducing 
a  short  link  BD,  Fig.  402.  Thus,  if,  in  DJi  produced,  BG  be 
taken  to  represent  the  resultant  stress  along  the  link,  and  if 
the  parallelogram  HK  be  completed,  J)K  will  represent  the 
stress  along  JiJi,  and  JUI  that  along  Jil''. 

This  lini.  device  has  been  employed  to  equalize  the  pressure 
on  the  turn-table  TT  of  a  swing-bridge  (Fig.  403).  An  "  equal- 
izer" or  "  rocker-link"  BD,  Fig.  404,  conveys  the  stresses  trans- 
niit*^"'  *^hrough  the  members  of  the  truss  terminating  in  J)  to 
tht   Ci'MVe  posts  J)T. 

'1  heoretically,  therefore,  the  pressure  over  TT  will  be  evenly 
distributed,  whatever  the  loading  may  be,  if  the  direction  of 
HI)  bisects  the  angle  TB'f  and  if  friction  is  neglected. 

The  joint  between  the  central  span  and  the  cantilever  re- 
quires the  most  careful  consideration  and  should  fulfil  the 
tollinving  conditions: 

(a)  The  two  cantilevers  should  be  free  to  expand  and  con- 
tract under  changes  of  temperature. 

{/>)  Tlie  central  span  should  have  a  longitudinal  support 
which  will  enable  it  to  withstand  the  effect  of  the  braking  of  a 
train  or  the  pressure  of  a  wind  blowing  longitudinally. 

(f)  The  wind-pressure  on  the  central  span  should  bear 
equally  on  the  two  cantilevers. 

{i/}  The  connections  at  both  ends  should  have  sufficient 
lateral  rigidity  to  check  undue  lateral  vibration.  Conditions 
Ui)  and  {c)  would  be  fulfilled  by  supporting  the  central  span 
like  an  ordinary  bridge-truss  upon  a  rocker  bolted  down  at  one 
ciul  and  upon  a  rocker  resting  on  expansion  rollers  at  the 
other.  This,  however,  would  not  satisfy  condition  (/;).  It  is 
preferable  to  support  the  span  by  means  of  rollers  or  links  at 
botii  ends,  and  to  secure  it  to  one  cantilever  only  on  the 
ccntr.d  line  of  the  bridge  with  a  large  vertical   pin,  adapted  to 


630 


THEORY  OF  STRUCTURES. 


transmit  all  the  lateral  shearing  force.  A  similar  pin  at  the 
other  end,  free  to  move  in  an  elongated  hole,  or  some  equiva- 
lent arrangement,  as,  e.g.,  a  sleeve-joint  bearing  laterally  and 
with  rollers  in  the  seat,  is  a  satisfactory  method  of  transmitting 
the  shearing  force  at  that  end  also.  (If  there  is  an  end  post,  it 
may  be  made  to  act  like  a  hinge  so  as  to  allow  for  expansion, 
etc.)  The  points  of  contrary  flexure  of  the  whole  bridge  under 
wind-pressure  are  thus  fixed,  and  all  uncertainty  as  to  wind- 
stresses  removed. 

Where  other  spans  have  to  be  built  adjacent  to  a  large  can- 
tilever span,  it  should  not  be  hastily  assumed  that  it  is  neces- 
sarily best  to  counterbalance  the  cantilever  by  a  contiguous 
cantilever  in  the  opposite  direction.  If  it  is  possible  to  obtain 
good  foundations  and  if  piers  are  not  expensive,  it  might  be 
cheaper  to  build  a  number  of  short  independent  side  spans  and 
to  secure  the  cantilever  to  an  independent  anchorage.  If  this 
is  done,  care  must  be  taken  to  give  the  abutiiunt  sufficient  sta- 
bility to  take  up  the  unbalanced  thrust  along  the  lower  boom 
of  the  cantilever. 

Suppose  that  the  cantilever  is  anchored  back  by  means  of 
a  single  back-stay. 

Let  W  =■  weight  necessary  to  resist  the  pull  of  the  back- 
stay ; 
h  =  depth  of  end  post  of  cantilever ; 
z  =  horizontal   distance    between    foot   of  post  and 
anchorage ; 
M  —  bending  moment  at  abutment  =  Wz. 

If  it  is  now  assumed  that  the  sectional  areas  of  the  post 
and  back-stay  are  proportioned  to  the  stresses  they  have  to  bear 
(which  is  never  the  case  in  practice),  the  quantity  of  material  in 
these  members  must  be  proportional  to 


h 


hs 


which  is  a  minimum  when  z  =  s/2h. 

If  a  horizontal  member  is  introduced  between  the  feet  of 


CA  .V I  'IL  E I  'ER    y  'A'  i  \SSE  S. 


6'M 


the  back-stay  and  the  post,  the  quantity  of  material  becomes 
proportional  to 

W"-±-~  +  W/i  +  IV^-  =  2M''  ^" 


k 


h 


zli 


which  is  a  minimum  when  z  =  h,  i.e.,  when  the  back-stay  slopes 
at  an  an<Tle  of  45°.  By  making  the  angle  between  the  back- 
stay and  the  horizontal  a  little  less  than  45'',  a  certain  amount 
of  material  may  be  saved  in  the  joints  of  the  back-stays  and 
also  in  the  anchors,  which  more  than  compensates  for  the  in- 
creased weight  of  the  anchors  themselves. 

(Note. — In  these  calculations  it  is  also  assumed  that  the  top 
chord  is  horizontal,  and  that  the  feet  of  the  post  and  back  stay 
arc  in  the  same  horizontal  plane.  This  is  rarely  the  case  in 
practice.) 

According  to  the  above  the  weight  of  material  necessary 
for  the  back-stay  xs  directly  proportional  to  the  bending  moment 
at  the  abutment  and  ///TYTi^/j' proportional  to  the  depth  of  the 
cantilever,  other  things  being  equal.  A  double  cantilever  has, 
'11  general,  no  anchorage  of  any  great  importance. 

If  the  span  is  very  great,  a  cantilever  bridge  usually  re- 
quires less  material  tiian  any  other  rigid  structure  of  equal 
strength,  even  though  anchorage  may  have  to  be  provided. 
If  two  large  spans  are  to  be  built,  a  double  cantilever,  requir- 
ing no  anchorage,  may  effect  a  very  considerable  saving  in 
material,  although  a  double  pier,  of  sufficient  width  for  stability 
under  all  conditions  of  loading,  will  be  necessary. 

Again,  where  false-works  are  costl\-  or  impossible,  tlie 
juoperty  of  the  cantilever,  that  it  can  be  made  to  support 
itself  durinj':  erection,  gives  it  an  immense  advantage.  If  the 
design  of  the  cantilever  is  such  that  it  can  be  built  out  rapidly 
and  cheaply,  it  will  often  be  the  most  economical  frame  in  the 
ond,  even  if  the  total  ([uantity  of  material  is  not  so  small  as 
*hat  required  for  some  other  type  of  bridge.  In  all  engineering 
work",  quantity  Oj  material  is  only  one  of  the  elements  of  cost, 
and  this  should  be  carefully  borne  in   mind  when  designing  a 

itilever  bridge,  because  a  want  of  regard  to  the  method  of 


632 


THEORY  OF  STRUCTURES. 


erection  may  easily  add  to  its  cost  an  amount  much  greater 
than  can  be  saved  by  economizing  material. 

In  ordinary  bridge-trusses  the  amount  of  the  web  metal  iy 
greatest  at  the  ends  and  least  at  the  centre,  while  the  amount 
of  the  chord  meial  is  least  at  the  ends  and  greatest  at  t':c 
H:entre.  Thus,  the  assumption  of  a  uniformly  distributed  dead 
load  for  such  bridges  is,  generally  speaking,  sufficiently  ac- 
curate for  practical  purposes.  In  the  case  of  cantilever 
bridges,  however,  the  circumstances  are  entirely  different.  In 
these  the  amount  of  the  metal  both  in  the  web  and  in  the 
chords  is  greatest  at  the  support  and  least  at  the  end.  For 
example,  the  weight  of  the  cantilevers  (exclusive  of  the  weigh.t 
of  platform,  viz.,  ^  ton  per  lineal  foot)  for  the  Indus  Bridge, 
per  lineal  foot,  varies  from  6^  tons  at  the  supports  to  i  ton  at 
the  outer  ends.  Hence,  the  lu'pothesis  of  a  uniformly  dis- 
tributed dead  load  for  such  structures  cannot  hold  good. 

The  weight  of  a  cantilever  for  a  given  span  may  be  approxi- 
mately calculated  in  the  following  manner : 

Determine  the  stresses  in  the  several  members,  panel  by 
panel — 

(A)  For  a  load  consisting  of 

(i)  a  given  unit  weight,  say  lOO  tons,  at  the  outer  end  ; 
(2)  the  corresponding  dead  weight. 

(B)  For  a  load  consisting  of 

(1)  the  specified  live  load  ; 

(2)  the  corresponding  panel  dead  weight. 

Thus,  the  whole  weight  of  a  panel  will  be  the  sum  of  the 
weights  deduced  in  (Ai  and  (B),  and  the  total  weight  of  the 
cantilever  will  be  the  sum  of  the  several  panel  weights. 

This  process  evidently  gives  at  the  same  time  the  weitjht:^ 
of  cantilevers  of  one,  two,  three,  etc.,  panel  lengths,  the  load- 
remaining  the  same. 

The  panel  dead  weights  referred  to  in  (A)  and  (B)  must,  in 
the  first  place,  be  assumed.  This  can  be  done  with  a  large  de- 
gree of  accuracy,  as  the  dead  weight  must  necessarily  ^'■;v7(^/W/r 
increase  towards  the  support,  and  any  error  in  a  particular 
panel  may  be  easily  rectified  by  subsequent  calculations. 


w 


L>  J'!'  ' 


CAN  TILE  VER    TR  USSES. 


633 


Agali,  the  preceding  remarks  iiulicatc  a  method  of  finding 
the  most  economical  cantilever  length  in  any  given  case. 

Tak  J,  e.g.,  an  opening  spanned  by  two  equal  cantilevt-rs  and 
t\\  int'jrmediatc  girder.  Having  selected  the  type  of  bridge  to 
be  cuipioyed  for  the  intermediate  span,  estimate,  either  from 
existing  bridges  or  otherwise,  the  weights  of  independent 
budges  of  the  same  type  and  of  different  spans.  Sketch  a 
skeleton  diagram  of  the  cantilever,  extending  over  one-half  of 
the  whole  span,  and  apply  to  it  the  processes  referred  to  in  (A) 
and  \\\\ 

If  /.is  the  length  of  the  cantilever  and  P  that  of  a  panel, 
the  following  table,  in  which  the  intermediate  span  increases 
by  two  panel  lengths  at  a  time,  may  be  prepared  : 


c  _ 

2S 

End 

lever 
nier- 

c 
U 

Can- 
for 
tons 
[i. 

Can- 
lie 
at 
rom 
diate 

Can- 

ue  to 
ILive 
d  its 
ight. 

£il 

0.-* 

•0 

■3  B   . 

0  u 

"-§s 

"-UH 

&s 

U^  u  — 
'J^  «  v  rt 

*i  U  C  C  I-  c 

3  s 
^•^  0. 

- 

X 

^ 

IS 

^ 

H 

0 

L 

1P 

L 

-  2P 

4/" 

L 

-aP 

bl' 

L 

-  6P 

%P 

L 

-  8/" 

etc. 

etc. 

Weight  in  col.  3  =  one-half  oi  the  weight  of  the  intermediate 

girder 
■\-  one-half  of  the  live  load  it  carries  if  uni- 
formly distributed.  (The  proportion  will 
be  greater  than  one-half  for  arbitrarily 
distributed  loads,  and  may  be  easily  de- 
termined in  the  usual  manner.) 
Col.  5  gives  the  weights  obtained  as  in  A. 

weight  on  end  of  cantilever 

Col.  0  =  col.  5  X ■ . 

^  100 

Col.  7  gives  the  weights  obtained  as  in  B, 

Col.  8  =  col.  2  +  col.  6  -|-  col.  7. 

It  is  iirportant  to  bear  in  mind  that  an  increase  in  the  weight 

of  the  central  span  necessitates  a  corresponding  increase  in  the 


11 


^34 


THEORY  OF  STRUCTURES. 


weights  of  the  cantilevers.  Hence,  in  order  that  the  weight  of 
the  structure  may  be  a  minimum,  the  best  material  with  the 
highest  practicable  working  unit  stress  should  be  employed  for 
the  centre  span. 

The  table  must  of  course  be  modified  to  meet  the  require- 
ments of  different  sites.  Thus,  if  anchorage  is  needed,  a  column 
may  be  added  for  the  weights  of  the  back-stays,  etc. 

i6.  Curve  of  Cantilever  Boom. — Consider  a  cantilever 
with  one  horizontal  boom  OA,  and  let  x,  y  be  the  co-ordinates 
of  any  point  P  in  the  other  boom,  O  being  the  origin  of  co-or- 

i 


— * 


f-3* 


405- 


Fig.  406. 

dinates  and  A  the  abutment  end  of  the  cantilever. 

Let  Whit  the  portion  of  the  weight  of  an  independent  span 

supported  at  O. 

Let  w  be  the  intensity  of  the  load  at  the  vertical  section 

through  P. 

Assume  (i)  that  there  are  no  diagonal  strains,  and,  hence, 
that  the  web  consists  of  vertical  members  only; 

(2)  that  the  stress  H  in  the  horizontal  boom  is 

constant,  and  therefore  the  bending  moment 
^tP=Hy\ 

(3)  that  the  whole  load  is  transmitted  through  the 

vertical  members  of  the  web. 

Let  k  be  such  a  factor  that  kTl  is  the  weight  of  a  member 
of  length  /,  subjected  to  a  stress  T. 

{Note. — If  /  is  in  feet  and  T  in  tons,  then  k  for  steel  is  about 
.0003,  allowance  being  made  for  loss  of  section  or  increase  of 
weight  at  connections.) 

w  consists  of  two  parts,  viz.,  a  constant  part  p,  due  to  the 
weight  of  the  platform,  wind-bracing,  etc.,  which  is  assumed  to 


El  cantilever 
co-ordinates 
:Tin  of  co-or- 


CURVE   OF   CANTILEVER  BOOM. 


635 


be   uniformly  distributed ;  and    a    variable  part,  due    to    the 
weight  of  the  cantilever,  which  may  be  obtained  as  follows: 
Weight  of  element  dx  of  horizontal  boom  =  kHdx, 

"    web  corresponding  to  dx  ~  kwydx. 

"   element  of  curved  boom  corresponding   to  dx 


« 


=./.0^)'... 


Hence  the  variable  intensity  of  weight 

ds' 


=  kH+kwy  +  kH[£)\ 


and 


ds\' 


w  =p-\-  kH-\-  kwy -(- kH  \-j-\ 


Again,  if  M  is  the  bending  moment  and  S  the  shearing 
force  at  the  vertical  section  through  P,  then 

d'M  _dS_      _      dy 

dx'   ~  dx  ~      ~       dx' ' 


.'.H^j^,=p  +  kH-\-kwy  +  kH  (~J 


dx' 


Integrating  twice. 


Iiy  =  A-\-Bx  +  {p+2kH)^+kH^, 


A  and  B  being  constants  of  integration. 


When  X  —  o,    y  =  o,     and     H^  ~  W. 


.dy 
dx 


Thus, 


^  =  o     and  B=  W. 


636 


Hence, 


THEORY  OF  STRUCTURES. 


Hy  =  Wx  +  (/.  +  2kHf~  +  kH^ 


is  the  equation  to  the  curve  of  the  boom,  and  represents  an 
eUipse  with  its  major  axis  vertical,  and  with  the  lengths  of  the 

two  axes  in  a  ratio  equal  to  I — t^t j  • 

The  depth  of  the  longest  cantilever  is  determined  by  the 
vertical  tangent  at  the  end  of  the  minor  axis,  and  corrcsijoiids 

to  the  value  of  y  given   by  making  — -  =  o  m  the  preceding 

equation,  which  gives  y  =  — . 

For  a  given  value  of  H  the  curve  of  the  boom  is  independ- 
ent of  the  span.  Again,  for  a  given  length  of  cantilever  with 
a  boom  of  this  elliptic  form,  a  value  of  H  may  be  found  which 
will  make  the  total  weight  a  minimum,  and  which  will  there- 
fore give  the  most  economical  depth.  Such  an  investigation, 
however,  can  only  be  of  interest  to  mathematicians,  as  the 
hypotheses  are  far  from  being  even  approximately  true  in 
practice,  and  the  resulting  depth  would  be  obviously  too  great. 

Assumption  (l)  on  page  634  no  longer  holds  when  a  live 
load  has  to  be  considered.  Diagonal  bracings  must  then  be 
introduced,  which  become  heavier  as  the  depth  increases,  in 
consequence  of  their  increased  length.  The  diagonal  bracings 
are  also  largely  affected  by  the  length  of  the  panels.  If  the 
panels  are  short,  and  if  a  great  depth  of  cantilever,  diminishing 
rapidly  away  from  the  abutment,  is  used,  the  angles  of  the 
diagonal  bracing,  near  the  abutment,  will  be  unfavorable  to 
economy.  This  difficulty  may  be  avoided  by  adopting  a 
double  system  of  triangulation  over  the  deeper  part  of  the 
cantilever  only,  or  even  a  treble  system  for  some  distance  in 
a  large  span.  The  objections  justly  urged  against  multiple 
systems  of  triangulation  in  trusses  lose  most  of  their  force  in 
large  cantilevers.  In  the  first  place,  the  method  of  erection 
by  building  out  injures  that  each  diagonal  shall  take  its  proper 
share  of  the  dead  load ;  and  in  the  second  place,  it  should  be 


mmw^ 


CURVE   OF  CANTILEVER  BOOM. 


637 


remembered  that  only  in  large  spans  c(jLild  a  double  system 
have  anytliin;^  to  reeommend  it,  and  then  only  near  the  abut- 
ment where  the  stresses  are  greatest :  in  sueh  cases  the  moving 
load  only  produces  a  small  portion  of  the  entire  stress  in  the 
web.  In  practice,  a  compromise  has  to  be  made  between  dif- 
tcrent  requirements,  and  the  depth  must  be  kept  within  such 
limits  as  will  admit  of  reasonable  proportions  in  other  respects, 
while  the  diagonal  ties  or  struts  may  be  allowed  to  vary  in  in- 
clination, to  some  extent,  from  one  panel  to  another. 

Again,  in  h.King  the  panel  length,  care  must  be  taken  that 
there  is  no  undue  excess  of  platform  weight,  as  this  will  pro- 
duce a  corresponding  increase  in  the  weight  of  the  cantilever. 

An  excessive  depth  of  cantilever  generally  causes  an  in- 
crease in  the  cost  of  erection. 

Both  theory  and  practice,  however,  indicate  that  it  will  be 
more  advantageous  to  choose  a  greater  depth  for  a  cantilever 
than  for  an  ordinary  girder  bridge. 

An  ordinary  proportion  for  a  large  girder  bridge  would  be 
one-ninth  to  one  seventh  of  the  span,  and  if  for  the  girder  were 
substituted  two  cantilevers  meeting  in  the  middle  of  the  span, 
the  depth  might  with  advantage  be  considerably  increased 
beyond  this  proportion  at  the  abutment,  if  it  be  reduced  to  )iil 
where  the  cantilevers  meet.  When  a  central  span  is  introduced, 
resting  upon  the  ends  of  the  two  cantilevers,  the  concentrated 
load  on  the  end  gives  an  additional  reason  for  still  further  in- 
creasing the  depth  at  the  ■A\iw\.xt\Q\\X.  proportionally  to  the  Ungtii 
of  the  cantilever.  The  greatest  economical  depth  has  probably 
been  reached  in  the  Indus  bridge,  in  which  the  depth  at  the 
abutment  =  ,54  X  length  of  cantilever.  Probably  the  propor- 
tion of  one-third  of  the  length  of  the  cantilever  would  be 
ample,  except  where  the  anchorage  causes  a  considerable  part 
of  the  whole  weight,  but  each  case  must  be  considered  on  its 
own  merits.  The  reduction  of  deflection  obtained  by  increas- 
'vAs^  the  depth  is  also  an  appreciable  consideration. 

If  a  depth  be  chosen  not  widely  different  from  that  which 
makes  the  quantity  of  material  a  minimum,  the  weight  will  be 
only  slightly  increased,  while  it  is  possible  that  great  structural 
advantages  may  be  gained  in  other  directions.     In  recommend- 


ii 


.,   if 


IMAGE  EVALUATION 
TEST  TARGET  (MT-3) 


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Corporation 


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638 


THEORY  OF  STRUCTURES. 


ing  a  great  depth  for  a  cantilever  at  its  abutment,  it  is  assumed 
that  the  depth  will  be  continuously  reduced  from  the  abutmcMit 
outwards.  If  the  load  were  continuously  distributed,  it  is  by 
no  means  certain  that  a  cantilever  of  uniform  depth  would  re- 
quire  more  material  than  one  of  varying  depth,  but  it  has 
already  been  pointed  out  to  what  extent  the  weight  of  the 
structure  itself  necessarily  varies,  and  if  the  concentrated  load 
at  the  end  were  separately  considered,  the  economical  truss 
would  be  a  simple  triangular  frame  of  very  great  depth.  From 
economic  considerations,  it  would  be  well  to  reduce  the  depth 
of  the  cantilever  at  the  outer  end  to  nil,  but  in  many  cases  it 
is  thought  advisable  to  maintain  a  depth  at  this  point  equal  to 
that  at  the  end  of  the  central  span,  so  that  the  latter  may  be 
built  out  without  false-works,  under  the  same  system  of  erection 
as  is  pursued  in  the  case  of  the  cantilever.  The  post  at  the 
ends  of  the  central  span  and  cantilever  is  sometimes  hinged  to 
allow  for  expansion. 

17.  Deflection. — A  serious  objection  urged  against  can- 
tilever bridges  is  the  excessive  and  irregular  deflection  to  whicli 
they  are  sometimes  subject.  They  usually  deflect  more  than 
ordinary  truss-bridges,  and  the  deflection  is  proportionately 
increased  under  suddenly  applied  loads.  In  the  endeavor  to 
lecover  its  normal  position,  the  cantilever  springs  back  with 
increased  force  and,  owing  to  the  small  resistance  offered  by 
the  weight  and  stiffness  at  the  outer  end,  there  may  result, 
especially  in  light  bridges,  a  kicking  movement.  It  must,  how- 
ever, be  borne  in  mind  that  the  deflection,  of  which  the  impor- 
tance in  connection  with  iron  bridges  has  always  been  rcco<f- 
nized,  is  not  in  itself  necessarily  an  evil,  except  in  so  far  as  it  is 
an  indication  or  a  cause  of  over-strain. 

18.  The  Statical  Deflection,  due  to  a  quiescent  lo.u! 
must  be  distinguished  from  what  might  be  called  the  dynamical 
deflection,  i.e.,  the  additional  deflection  due  to  a  load  in  motion. 
The  former  should  not  exceed  the  deflection  corresponding  to 
the  statical  stresses  for  which  the  bridge  is  designed.  '1  lie 
amount  of  the  dynamical  deflection  depends  both  upon  the 
nature  of  the  loads  and  upon  the  manner  in  which  they  aic 
applied,  nor  are  there  sufficient  data  to  determine  its  value 


LIVE  LOAD. 


639 


even  approximately.     It  certainly  largely  increases  the  statical 
stresses  and  produces  other  ill  effects  of  which  little  is  known. 

Hitherto,  the  question  as  to  the  deflection  of  framed  struc- 
tures has  received  but  meagre  attention,  and  formulae  deduced 
for  solid  girders  have  been  employed  with  misleading  results. 
It  would  seem  to  be  more  scientific  and  correct  to  treat  each 
member  separately  and  to  consider  its  individual  deformation. 

19.  Rollers. — One  end  of  a  bridge  usually  rests  upon  nests 
of  turned  wrought-iron  or  steel  friction  rollers  running  between 
planed  surfaces.  The  diarrieter  of  a  roller  should  not  be  less 
than  2  inches,  and  the  pressure  upon  it  in  pounds  per  lineal 
incli  should  not  exceed  5CXD  ^d  ii  made  of  wrought-iron,  or 
600  \^d\[  made  of  steel,  d  being  the  diameter  in  inches. 

20.  Live  Load. — It  is  a  common  practice  with  many  en- 
gineers to  specify  the  live  load  for  a  bridge  as  consisting  01  a 
number  of  arbitrary  concentrated  weights  which  are  more  or 
less  equivalent  to  the  loads  thrown  upon  the  locomotive  and 
car  axles. 

Figs.  407,  408,  and  409  are  examples  of  such  practice. 


"^mm 

i'ii^)^^ 

■ 

■"'"**' 

1 

■ 

till 

\ 

y^mmKjhm 

II 

I 

64  ei 


Fig. 


407. 


/(7)    w-  G)'^G)'(*)G)  »*•(?)  r.8-C7).»-(7)»a-C7)  ^/(^  \ 


M  CI 


Fig.  408. 


v,l 


I      I     i   i    i  •  i    i     i   I        I 

Fig.  409. 


With  such  a  live  load,  the  determination  of  the  position  of 
the  locomotive  and  cars  which  will  give  a  maximum  shear  and 
a  maximum  bending  moment  at  any  section  is  much  facilitated 
by  the  principles  enunciated  in  Art.  8,  Chap.  II. 


;;■  I! 


640 


THFORY  OF  STRUCTURES. 


If  the  chords  are  parallel,  and  if  5  is  the  maximum  shear 
transmitted  through  a  diagonal  inclined  at  an  angle  B  to  the 
vertical,  the  maximum  stress  in  that  diagonal  =  5  sec  ^,  and 
the  corresponding  stress  transmitted  to  a  chord  through  the 
diagonal 

=  5  sec  ^  sin  ft  =  S  tan  0. 

A  modification  is  necessary  when  the  chords  are  not  paral- 
lei.  Consider,  e.g.,  a  truss  with  a  horizontal  bottom  chord 
and  a  top  chord  composed  of  inclined  members.  Retain  the 
same  notation  as  in  the  article  referred  to,  and  let  D^ ,  D^  be 
the  stresses  corresponding  to  the  Jirsl  and  .y^^ow^  distributions, 
respectively,  in  a  diagonal  met  by  a  vertical  section  between 
the  rth  and  {r -\-  i)th  weights.  Also,  let  the  member  of  the 
upper  chord  cut  by  the  same  section  be  produced  to  meet  the 
horizontal  chord  produced  in  the  point  C. 


Let  AC  —  h,  and  let  p  be  the  perpendicular  from  C  upon 
the  diagonal  in  question. 
Taking  moments  about  C, 


D,p  -  RJi  -  w^h  +  /  —  a,)  —  "^l^t  +  ^  -  «,) 


and 


Wr{h  -\- 1  —  a,\ 


Dj>  —  RJt  —  w,{h  -\- 1  —  a,  —  x)  —  w^h  -\-l  —  a^  —  x)  —  . . . 

—  Wjjl  -\-l—ar  —  x)  —  7CV+,(/^  -\-l—  (Ir^x  —  x)  —  ... 

—  Wr+lh  -\-l—  a,+^  -  x). 

It  is  assumed,  for  simplicity,  that  no  weights  leave  or  ad- 
vance upon  the  bridge.  '  ' 


r,-  ,,--. 


.-.  A  =  A, 


LIVE  LOAD. 


641 


according  as 


RJi  -  will  -f  /  -  «^)  ^  wih  +  /  -  ^,)  -  .  .  .  -_  w,{h  +  l-a^)  = 

RJi  —  w,{h  +  /—«,  —  ■*')  —  ^^>('''  +  /  —  <7,  —  ^)  —  .  . . 

—  wXh  +  /  — «r  —  ^)  —  Wr+,(/'  +  /  —  <?,+,  —;«:)—... 

—  Z£',+,(/!  +  /  -  ^,+,  4-  x), 
or 

^/(^  +  ^0   =  -^'(^l  +«'.  +  •••  +  W'r  +  Wr+.  +  •  •  •  Wr+j) 

where  Rg\l-\-  h)  =  algebraic  sum  of  the  moments,  with  respect 
to  C,  of  the   weights  transferred, 

=  ^i>r+,{h  +  /  -  «.+,)  +  .  ■  .  +  w,+,(A  4-  /  —  a^^.,) 
and 


Hence, 


according  as 


R,-R,=jW^. 


A I  A. 


R:{l-^h)^x{W^-\-T)+W„jh. 


Take,  e.g.,  the  truss  represented  by  the  accompanying  dia- 
ijrain  (Sault  Ste.  Marie  Bridge),  the  live  load  being  that  shown 
by  Fig.  411,  i.e.,  the  loading  from  a  Standard  Consolidation 
engine  with  four  drivers  and  one  leading  wheel. 


.lOrtJt 


\  cnite  1 


X  IM         •sj'*        •^  M^         M  J  ^  'X  'J 

A"rr  '-P,""" '  p,"*  W','""  •P4""  'p,     i^     H     ^     Pi 


Fig.  411. 

Span  =  239  ft. 

Length  of  centre  verticals  =  40  ft.^  of  end  verticals  =  27  ft. 


■f  » 


642 


THEORY  OF  STRUCTURES. 


Applying  the  principles  referred  to  in  the  preceding  it  is 
found  that  the  distributions  of  live  load,  concentrated  at  tin 
panel  points,  which  will  give  the  maximum  stresses  in  the 
several  members,  may  be  tabulated  as  below : 


Distribu- 
tions, 


Case  I  . 

a  . 
3' 
4  • 
5- 
6. 

7  • 
8. 


Dead  weight 


End 

Reac- 

Load 

Load 

Load 

Load 

Load 

Load 

Load 

tion 
MA. 

at/,. 

at/j. 

at/,. 

at/4. 

at/,. 

at/,. 

at/,. 

•87990 

49500 

38700 

459»5 

43750 

36225 

36000 

36000 

162920 

49500 

38700 

45925 

43750 

36225 

36000 

124230 

6400 

47200 

40200 

43400 

45800 

37100 

9S030 

6400 

47200 

40200 

43400 

45800 

69410 

6400 

47200 

40J00 

43400 

47400 

6400 

47200 

40200 

29100 

6400 

47200 

15380 

6400 

121500 

27000 

27000 

27000 

27000 

27000 

27000 

27000 

Load 
at/,. 


36000 
36000 
36000 
37100 
45800 
43400 
40200 
47200 
27000 


Load 
at/,. 


360C0 
3<iooo 
36000 
360C0 
37"oo 
45800 
4.1400 
40200 
27000 


N.B. — These  numbers  are  convenient  whole  numbers  within 
about  one-half  of  one  per  cent  of  the  calculated  results.  The 
panel  length  is  also  assumed  to  be  24  ft. 

In  Cases  i  and  2  the  //«></ driver  is  at  a  panel  point ;  in  the 
remaining  cases  the  second  driver  is  at  a  panel  point. 

The  dead  weight  includes  the  weight  of  the  ironwork  and 
flooring.  The  panel  loads  may  be  easily  calculated,  either 
analytically  or  graphically.  For  example,  let  A,  B,  C,  D  he 
four  consecutive  panel  points,  and  let  the  t/iird  driver  be  at  B. 

Panel  load  at  A 

198   ,  /108  +  52\  ^  „ 

=    750o-|g  +  12000^ — i^8"^/  =  "^^3.  say  11,900  lbs. 

Panel  load  at  B 

90    ,             /i8o+236-f-288+232\       ,^,J]4i±Z7 
=    75002^+i20oo( ^gg l  +  '^'S[-m-J 

=  49387,  say  49,500  lbs. 

Panel  load  at  C 

{S6-\-2H\  ,      ,     /i87-f-3ii+284iH-22oi\  118^ 

=  '200o(^-58^)+^°^^5(~^ ^88 i+7500^, 

=  38445,  say  38,700  lbs.  . 


LIVE  LOAD. 


643 


% 


Or,  graphically,  upon  the  vertical  through  B  (Fig.  412)  take 
BM  to  represent  7500  lbs.,  and  join  AM.  Let  the  vertical 
through  rt,  meet  AM  in  ^, ,  and  the  horizontal  through  AM 
iiw,.  Then  aj}^  represents  the  portion  of  7500  lbs.  borne  at 
/)',  and  b^c^  the  portion  borne  at  A. 

Also,  take  BN  to  represent  12,000  lbs. ;  join  AN,  CN.  Let 
the  verticals  through  a,,  a^,  a^  meet  AN,  CN  in  b^,  b^,  b^,  and 
the  horizontal  through  N  in  t,,  c^,  c^.     Then  a,^,,  a/,,  a«d« 


represent  the  portions  of  each  12,000  lbs.  borne  at  .5,  while 
/ij^,  b^c^  represent  the  portions  borne  at  A,  and  b^c^  the  portion 
borne  at  C. 

Finally,  take  BO  to  represent  10,625  lbs.,  and  join  CO.  Let 
the  verticals  through  a^,  a^  meet  CO  in  b^,  b,,  and  the  horizon- 
tal thro'jgh  O  in  c^,  Cf.  Then  a^b^,  a^b,  are  the  portions  of  each 
10.625  lbs.  borne  at  B,  while  b^c^,  b^c,  are  the  portions  borne  at 
C".    Thus  the  total  weight  at  B 

=  «/.  +  a  A  +  «A  +  W+  ^A  -t-  a  A  +  «/.• 

It  is  open  to  grave  question  whether  the  extremely  nice 
calculations  required  by  the  assumption  of  arbitrary  weight 
calculations  are  not  unnecessary  except  for  floor  sy.stems.  The 
constantly  increasing  locomotive  and  car  weights  and  the 
variety  in  type  of  locomotive  would  seem  to  render  such  cal- 
culations, based  as  they  are  upon  one  particular  distribution  of 
ioad,  of  no  effect. 

On  the  other  hand,  if  it  is  assumed  that  the  standard  live 
load  consists  of  a  uniform  load  of,  say,  3000  lbs.  to  3600  lbs. 
per  lineal  foot,  with  a  .single  weight  of,  say,  25,000  lbs.  to 
35,000  lbs.  for  each  tru.ss,  at  the  head  or  at  any  other  specified 


!, 


644 


THEORY  OF  STRUCTURES. 


point,  i.e.,  rolling  on  the  uniform  load,  the  calculations  woulii 
be  much  simplified  and  the  resulting  stresses  would  be  at  Iea^l 
as  approximately  accurate. 

Let  E  be  the  single  concentrated  load,  T  the  panel  train 
load,  and  D  the  panel  dead  load. 

Consider  a  truss  of  N  panels  with  a  single  diagonal  system, 
Fig.  413,  and  let  E  be  at  the  rth  panel  point. 


Fig.  413. 

The  shear  immediately  in  front  of  E  due  to  E 

the  shear  at  same  point  due  to  T 

__  T  {N-r-i){N-r) 
~  N  2  • 

the  shear  at  same  point  due  to  Z? 

_  D  N{N  -  2r  -f  I) 

~  iV  2 

Diagonal  Stresses. — The  maximum  diagonal  stresses  may 
now  be  easily  tabulated  as  follows : 

TABLE    I. 


I 


3 


N-1 


u 

a 
■o 

u 

3 

•a  . 

u  o 

>g 


<^-')l 


u  I 

"3 


N-i.N-i 


u 

3 


!1<T^ 


"5.- 

n 


A^- 


a.A^-i   T 

2  N 


11 
?*  > 

Bo 

S   •= 
— «'  s 

H 


'"  3  AT 


Q  V 


+ 

V. 
ct 

I 

a 

•a 
■3 


N.N— I 


■il 


52 


> 


!  S  ^ 

jj  — 

T  5 

-  "< 

•S  — 


N.N-iD 
2       A^ 


fifP 


LIVE  LOAD. 


645 


Col.  I  designates  the  several  diagonals. 

Col.  2  gives  the  multiplier  ^.V  —  r  for  different  values  of  r. 

Col.  3  gives  the  maximum  vertical  shears  due  to  E  trans- 
mitted through  the  several  diagonals.  This  shear  for  any 
given  diagonal  is  the  product  of  the  corresponding  multiplier 

in  col.  2  and  -jj . 

Cols.  4  and  5,  g  and  10  give  similar  quantities  for  the  live 
and  dead  loads. 

Col.  6  gives  the  sums  of  the  shears  in  cols.  3  and  5,  i.e.,  it 
gives  the  total  maximum  vertical  shears  due  to  live  load. 

Col.  8  gives  the  maximum  diagonal  stresses  due  to  live 
load.  For  any  specified  diagonal  it  is  the  product  of  the  cor- 
responding shear  in  col.  7  and  the  secant  of  the  angle  between 
the  vertical  and  the  diagonal  in  question. 

Col.  1 1  in  like  manner  gives  the  maximum  diagonal  stress 
due  to  dead  load. 

Col.  12  gives  the  total  maximum  diagonal  stresses  due  to 
both  live  and  dead  loads. 

Another  column  might  be  added  giving  the  sectional  areas 
of  the  diagonals. 

In  the  above  table  the  diagonal  stresses  due  to  the  live  and 
dead  loads  are  separately  determined,  as  different  coefficients 
of  strength  are  sometimes  specified  for  the  two  kinds  of  load. 
With  a  suitable  compound  coefficient  of  strength,  cols.  6,  8, 
and  II  may  be  replaced  by  a  column  giving  the  sums  of  the 
corresponding  shears  in  cols.  3,  5,  and  10.  These  sums,  multi- 
plied by  secant  ^,  give  the  maximum  diagonal  stresses. 

Stresses  in  the  Verticals. — The  maximum  stress  in  any  ver- 
tical, say  at  the  rth  panel  point,  is  evidently  the  vertical  com- 
ponent of  the  maximum  diagonal  stress  in  the  r\.\\  panel,  i.e.,  it 
is  the  maximum  vertical  shear  in  the  rth  panel. 

To  be  more  accurate,  this  amount  should  be  diminished  by 
the  portion  of  the  weight  of  the  lower  chord  borne  at  the  foot 
of  the  vertical  in  question. 

Chord   Stresses.  —  Take    the    load    at    each    panel    point 


t,  i'll  I i  1 


646 


THEORY  OF  STRUCTURES. 


TABLE  II.    (Compression  Chord.) 


TABLE 


% 

■8-6 

II 

■5 

c 

I 
■2 

un  fi 

c 

1 

5 

0   t 

Vertical 
transi 

c 

(4 

H 

Stress   tr 
to  Chor 
Diagon 

US 
"5  h 

13  3  456 

Col.  I  designates  the  chord  panel  length. 
Col.  3  gives  the  several  vertical  shears  transmitted  to  the 
chords   through   the    diagonals.      They   are   the    product   of 

-j^  \j^  -j-  ^  "f"  -^j  «^"^  ^^^  multipliers  in  col.  2. 

Col.  5  gives  the  chord  stresses  due  to  these  shears,  i.e.,  the 
product  of  the  shears  in  col.  3  and  the  corresponding  values  of 
tan  B  in  col.  4. 

Col.  6  gives  the  total  chord  stresses  in  the  several  panels. 
In  any  given  panel  the  total  chord  stress  is  equal  to  the  chord 
stress  due  to  the  shear  in  that  panel //?/j  the  total  chord  stress 
in  the  preceding  panel. 

Another  column  for  the  sectional  areas  of  the  several 
lengths  of  chord  may  be  added  if  required,  each  length  being 
designed  as  a  strut,  hinged  or  fixed  at  the  ends,  according  to 
^'iC  method  of  construction. 

A    precisely    similar 
table  may   be   prepared 
for  the  tension  chord. 
Example    i.    An 
'°  ■*''"■  «]^/i/-panelled  deck-truss 

of  108  ft.  span  and  18  ft.  deep,  with  a  single  diagonal  system. 

Concentrated  load  E  for  each  truss  =  25,000  lbs. 

Train  load  7"  for  each  truss  =  1600  lbs.  per  lineal  ft. 

=  2 1 ,600  lbs.  per  panel. 
Bridge  (dead)  load  D  for  each  truss  =  800  lbs.  per  lineal  ft. 

=  10,800  lbs.  per  panel. 
sec  0  =  \,     tan  6*  =  f . 


:« 

C 

0 

k' 

y 

7 
6 

5 
4 
3 


It  will 
mum  posU 
lbs.,  the  fo 
The  result 
l/iat  dtie  t 
minterbrac 
in  the  sixth 
in  order  to 

TABLE 


Chord  Si 


TABLE  OF 


Member. 


4(9 


mwi 


LIVE  LOAD. 


Un 


TABLE 

OF  MAXIMUM 

DIAGON.-XL   STRESSES.     (See  Table 

I.) 

"3 

c 

a 

k 
1 

00 

II 

li- 

1 

00 

"k 

1 

N 

i 

II 

loo 

567-X3 

Toul     Max. 
Shear  due  to 
Live  Load. 

• 
c 

1 

•fsi 
98218} 

t 

1 

» 

28 

Diag.     S'tress 
due  to  Dead 
Load. 

Total       Max. 
Uiag.  Stress. 

■^^ 

7 

21875 

21 

78575 

■  J 

37800 

4725" 

I454«l 

,/, 

6 

18750 

•5 

40500 

59250      li 

74062* 

30 

27000 

33750 

107812* 

'^. 

5 

15625 

10 

270fX) 

42625 

5328it 

12 

16200 

20250 

7353'i 

''f 

4 

12500 

6 

16200 

28700 

35875 

4 

5400 

0750 

42625 

</. 

3 

9375 

3 

8100 

•7475 

»>843l 

-   4 

-  5400 

-  6750 

"5093} 

i* 

a 

6250 

I 

2700 

8950 

11187* 

—  12 

—16200 

—20250 

•ii 

I 

3115 

3"5 

39o6i 

—  JO 

—27000 

-33750 

It  will  be  observed  that  in  the  fifth  panel  there  is  a  maxi- 
mum positive  shear  of  17,47S  'bs.  and  a  negative  shear  of  5400 
lbs.,  the  former  due  to  the  live  and  the  latter  to  the  dead  load. 
The  resultant  shear  of  12,075  lbs.,  which  is  opposite  in  kind  to 
that  due  to  the  dead  load,  is  provided  for  by  means  of  the 
counterbrace  ab.  No  counterbraces  are  theoretically  required 
in  the  sixth  and  seventh  panels,  but  they  are  often  introduced 
in  order  to  stiffen  the  truss. 

TABLE   OF  MAXIMUM   STRESSES  IK  THE  VERTICALS. 
^.  =  78575  +  37800  =  1 16.375  lbs. 
v^  —  59250  +  27000  =    86,250  " 
v^  =  27000  -|-  16200  =    43,200   " 
z/«=i62oo-j-    5400=    21,600  " 

Chord  Stresses. — Load  at  each  panel  point 


i 


=   g-+^+ ^  =  35.525  lbs. 


TABLE  OF   MAXIMUM   STRESSES   IN   COMPRESSION   CHORD. 


Member. 

4(9  -  ar). 

?'f^  =  44401. 

Tan  9. 

Chord  Stress  due 
to  Shear. 

Total  Max. 
Chord  Stress. 

Cl 

28 
20 
12 

4 

124337! 

888 1 2i 

53287! 
17762! 

93253! 
66609! 

39965* 
13321J 

93253! 
159862^ 
1998281 
213150 

648  THEORY  OF  STRUCTURES. 

TABLE  OF   MAXIMUM   STRESSES   IN   TENSION  CHORD. 


Member. 

4(9  -  ar). 

"r'=4.4o.. 

Tan  9. 

Chord  Stress  due       Total  M.i» 
to  Shear.            Chord  Stress 

t% 

28 
20 
12 

I24337i 
888i2i 

53287i 

J 
J 

30<j(J5l               »<W828J 

The  above  figures  may  be  checked  by  the  method  of 
moment.s. 

Note, — If  the  triKss  is  inverted  it  becomes  one  of  the  IIouc 
type.  The  stresses  are  the  same  in  magnitude,  but  reversed 
in  kind. 

Ex.  2.  An  eight-panel  through-bridge  of  the  double-inter- 
section  type  (Kig.  414^  iiav- 
ing    tlie    same    span,   depth, 
and  loading  as  in  Ex.  i. 
'^"2"'^^"^^^      i       '^      "  The    systems     1234... 

^'**-  ••'♦•  and     I   a  b  c  .  .  .    are     inde- 

pendent.    It  is  assumed  that  the  load  at  the  foot  of  the  end 
vertical  i^gh)  is  divided  equally  between  the  two  sy.stenis. 

TABLE   OF   MAXIMUM    STRESSES    IN  END-POSTS   AND 

DIAGONALS. 


is 
a 

Multiplier. 

10 
« 

m 

II 

.a 
0. 

"3 

S 

i 

II 
>0    00 

Total      Max. 
Shear  due  to 
Live  Load. 

u 

,)S 

Ma».       Diag. 
Stress  due  to 
Live  Load. 

i 

"a 

m 

II 

Diag.     Stress 
due  to  Dead 
Load. 

Tf.tal       Max.  ! 
Dia(j   Stress.   \ 

/ 

7 

21875 

21 

56700 

78575 

98218} 

28 

37800 

47350 

MM68J 

<i\ 

6 

18750 

6* 

>7S50 

36300 

I 

45375 

I2i 

16875 

21093} 

6646s} 

di 

S 

15625 

3i 

9450 

25075 

45>35 

«♦ 

"'475 

20655 

65790 

i* 

4 

12500 

a* 

6750 

1925° 

34650 

4i 

6075 

>oy35 

45585 

'',* 

3 

9375 

i 

1350 

10725 

«■ 

1930s 

i 

675 

•  2:5 

20520 

i* 

a 

6250 

i 

1350 

7600 

« 

13680 

-   3+ 

-   47*5 

-   8S05 

5"75 

i* 

k 

I56»* 

1562* 

I 

2812} 

-   7* 

— 10125 

-18325 

d-, 

k 

1562* 

1563* 

^ 

•953* 

-Hi 

-»55aS 

-27945 

The  counterbrace  cf  is  required  to  take  up  the  resultant 
shear  of  6250  —  4725  =  1525  lbs.,  which  is  opposite  in  kind  to 
that  due  to  the  dead  load. 

The  first  line  in  the  table  gives  the  maximum  thrust  alon^ 
the  end  post  (/).     It  is  made  up  of  the  stresses  transmitted 


m 


through  t 
at  the  firs 

TA 

The  m 
when  the  : 

V,  z 

Chords 


TABLE  OF 

Member       Multip 


2S 


Note. — c^  : 
TABLE  OF 


Membe 


Multi 
plier. 


Ex.  3.  A  t 
having  the  sam 


LIVE  LOAD. 


649 


throiii^h  the  two  systems  of  diagonals  when  the  2^,QOO  lbs.  is 
at  tlie  first  panel  point. 

TABLE   OF   MAXIMUM   STRESSES    IN  VERTICALS. 

The  maximum  stress  in  an  end  vertical  evidently  occurs 
when  the  25,000  lbs.  is  concentrated  at  its  foot. 

7',  =  25000  -f-  10800  =  35800  lbs.  (ten.sion) ; 
7',  =  19250-1-  6075  =  25325  "  (compression); 
7',=  10725 -j-   675  =  11400  "       " 
V,  =  6250—  4725  =  1525  " 

Chord  Stresses. — Load  at  each  panel  point 

=  |^-l-^-^-z)  =  35525. 


It 


TABLE  OF   MAXIMUM   STRESSES 

IN   COMPRESSION   CHORD. 

Member 

Multiplier. 

3"'5    -     A         ^ 

■  g      =  44-.0,. 

Tan  e. 

Chord  Siress 
due  tu  Shear. 

Tola! 

Maximum 

Chord 

Stress. 

(          28 

]      +I2i 

(     +    8i 

4i 

i 

1243371 

55507iJ 

37745t^s 

199821;! 

2220 ,»« 

* 
i 

93253i 

4163011* 

566171! 

•    i9i5oiSi 

29974  fl's 

3330JI 

22H76ji; 

22jSO()J 

Note. — c^  is  made  up  of  the  thrusts  transmitted  through 
TABLE  OF    MAXIMUM   STRESSES   IN   TENSION   CHORD. 


Member. 

Multi- 
plier. 

35535           . 

8      =444oi. 

TanO. 

Chord  Stress 
due  to  Shear. 

Total  Maximum 
Chord  Stress. 

28 
8i 

I24337i 
55507H 
37745  A 

i 

i 
f 

93253* 
4i630Ji 
5661 7|i 

93253i 
l348S3«f 

Ex.  3.  A  through-bridge   of  the  Warren  type  (Fig.  415) 
having  the  same  span  and  loading  as  in  Exs.  i  and  2. 

C\  Ct         C3         C4      


ti   ti  ti   ti 


liiil 


Hi 


n 


:> 

h 

1^^ 

i:-  ■; 

1 

1^ 

w 

t 

I: 

1 

i 

Fig.  415. 


€$0 


THEORY  OF  STRUCTURES. 


TABLE  OF   MAXIMUM   STRESSES   IN   DIAGONALS. 


2 

8 

■i 

Mazi 
hear 
itted. 

0 

E~. 

i 

a 

?3 

a 

3 
S 

1 

C4 

1 
00 

a 
21 

II 

NO     00 
N 

a, 

■3 
X 

II 
1  " 

V)  a 

las 

0  9  c« 

HBi: 

1-155 

3  0  . 
0  0  x 

■In 

d^=:d^ 

7 

2I87S 

56700 

28 

37800 

1 16375 

1 344 14 

d,=d. 

6 

.8750 

15 

J  0500 

20 

27000 

86250 

gi/)i() 

dt=.dt 

5 

15625 

ro 

■.'7000 

12 

16200 

58825 

<-'7i)43 

d,=dt 

4 

12500 

6 

16200 

4 

5400 

34100 

3<)386 

dt=dio 

3 

9375 

3 

8ioo 

-  4 

-    5400 

12075 

I3')47 

^ii=dii 

2 

6250 

I 

2700 

—  12 

—  16200 

dia=dn 

I 

3125 

—20 

—  27000 

The  resultant  stresses,  d,  =  </,„ ,  are  of  an  opposite  kind  to 
the  corresponding  stresses  due  to  the  dead  load.  Thus,  the 
diagonals  upon  which  they  act  must  be  designed  so  as  to  bear 
both  tensile  and  compressive  stresses.  The  stresses  </, ,  d^, 
//,,..;  are  compressions,  and  </,,  ^, ,</,,..  .  tensions. 


TABLE   OF   MAXIMUM   STRESSES    IN  COMPRESSION   CHORD. 

Mem- 
ber. 

Multi- 
plier. 

^-f^  =  4440*. 

Shear 
transmitted. 

Tanfi. 

i     Totiil 

Chord  Stress,  ^'^rrr 
1    Stress. 

j       28 
1       28 
j       20 
}        20 
(         12 
12 

■ 

4 
1        4 

Through  d,  124337* 
fl-,  124337* 
d»    88812* 
di    88812* 
'h    53287* 
"         dt    53287* 
d,    17762* 
dt    17762* 

248675 
■  177625 
[  106575 
•    35525 

•577 
•577 
•577 
•577 

143485  + 
102489  + 

61493  + 
20497  + 

143486 
245976 
307469 
327967 

TABLE   OF   MAXIMUM    STRESSES   IN    TENSION   CHORD. 

Mem- 
ber. 

Multi- 
plier. 

5f5  =  44.01. 

Shear 
transmitted. 

Tang. 

Chord  Stress. 

Total 

M<nxinium 

Chonl 

Stress. 

tt 
tt 

ti 

28 
28 
20 
20 
12 
12 

4 

Through  </,  124337* 

dt  124337* 
«/,    88812* 
dt    88812* 
rf,    53287* 
■  "         d,    53287* 
"         d,    177624 

124337* 
213150 

142100 

71050 

•577 
•577 

.577 

•577 

71742  + 
122987  + 

8199I  + 
40995  +  1 

71743 
19473' 

276722 

317718 

lymo  PXESSUKE. 


21-  Wind-pressure —N  "  ^St 

,""■"'  ""=  P'-^^ure  and   '..ioZ'TX    '''^"''"■^"ts    1„  detcr- 
yn,ea„s  of  feathers,  cloud     , ado  J      "'"''  ^"'  ''«"  ™ade 

]l"^  ^«ules,  either  through  errorfofK        '^""^  "«tr„me.-,ts 

Smeat,o„  inferred  from  RoL?'  "^  "bservations.  ^ 

«e  pressure  in  pounds  pe^sgu"  !  '^^'T'"'  ""'  ">=  aver. 
<"»"•)■  -^  200,  or  ^     "'^'"'"'  '«'  =  (velocity  in  „i,e,  p", 

200* 


According  to  Dines  the  ro....,3houId  be 


2000' 

,      The  Wind.Pressur*^  r^       .    . 

I  ,.™„,,  '«ure  Co„n,.ss,on  (Eng.)  recommended  the 


/>  = 


lOO' 


*o.t  ii  =  4  „ .    ,  "''*"'  «""d  velocities  should  be 

;-  ««,« JpttS'o/rstr  '""''^''  -  --ponding 

fc  tir  tr' ''---)  -id"her::ed°r  ^'^^  =■- 


€§2 


THEORY  OF  STRUCTURES. 


per  hour  would  be  13.1  lbs.  per  square  foot  according  to 
Smeaton's  rule  and  only  9.18  lbs.  according  to  Dines. 

Again,  certain  experiments  at  Greenwich  indicated  tliat 
the  pressure  was  increased  by  the  stiffness  of  the  copper  wire 
connecting  the  recording  pencil  with  the  pressure  plate,  and  a 
flexible  bra.  s  chain  was  therefore  substituted  for  the  wire. 
Thus  modified,  a  pressure  of  29  lbs.  per  square  foot  was  regis- 
tered as  corresponding  to  a  velocity  of  64  miles  per  hour. 
whereas  with  the  copper  wire  a  pressure  of  49!-  lbs.  per  .square 
foot  had  been  registered  with  a  velocity  of  only  53  mile.s  jxr 
hour. 

These  facts  tend  to  show  that  the  actual  pressure  is  much 
less  than  that  given  by  a  recording  instrument,  and  that  the 
very  high  pressures,  as,  e.g.,  80  lbs.  per  square  foot  and  even 
more,  must  be  due  to  gusts  or  squalls  having  a  purely  local 
effect.  This  opinion  seems  to  be  confirmed  by  Sir  B.  Baker's 
experiments  at  the  Forth  Bridge,  which  also  indicate  that  the 
pressure  per  square  foot  diminishes  as  the  area  acted  upon 
increases.  No  engineering  structure  could  withstand  a  press- 
ure of  80  lbs.  per  square  foot  of  surface,  and  a  pressure  of  28 
lbs.  to  32  lbs.  would  overturn  carriages,  drive  trains  from  the 
track,  and  stop  all  traffic. 

It  is,  of  course,  well  known  that  wind-forces  sufficiently 
powerful  to  uproot  huge  trees  and  to  demolish  the  strongest 
buildings  are  occasionally  developed  by  whirlwinds,  tornadoes, 
and  cyclones,  but  these  must  be  classed  as  acta  Dei  and  can 
scarcely  be  considered  by  an  engineer  in  his  calculations. 

Numerous  observations  as  to  the  effect  of  wind  upon  struc- 
tures in  different  localities  must  yet  be  made  before  any  useful 
and  reliable  rules  can  be  enunciated.  In  the  case  of  existing 
bridges  the  elongation  of  the  wind-braces  during  a  storm  can 
easily  be  measured  within  -^-^  of  an  inch.  Investigations 
should  be  made  as  to  the  action  of  the  wind  upon  surfaces  of 
different  forms  and  upon  sheltered  surfaces,  as,  e.g.,  upon  the 
surfaces  behind  the  windward  face  in  bridge-trusses.  Again, 
it  is  quite  possible,  if  not  probable,  that  many  of  the  recorded 
upsets  have  been  due  to  a  combined  lifting  and  side  action, 
requiring  a  much  less  flank-pressure  than  would  be  necessary 


EMPIRICAL  REGULATIONS. 


653 


if  there  were  no  upward  force,  and  hence  further  light  should 
be  obtained  on  this  point. 

Under  any  circumstances,  the  wind-stresses  should  be  as 
small  as  possible,  compatible  with  safety,  seeing  how  largely 
they  influence  the  sections  of  the  several  members,  especially 
in  bridges  of  long  span.- 

22.  Empirical  Regulations. 

Wind-Pressure  Coinmission  Rules. — For  railway  bridges  and 
viaducts  assume  a  maximum  pressure  of  56  lbs.  per  square  foot 
upon  an  area  to  be  estimated  as  follows  : 

A.  In  close-%\xdi&x  bridges  or  viaducts  the  area 

=  area  of  windward  face  of  girder 
-f-  area  of  train  surface  above  the  top  of  the  same  gir- 
der. 

B.  In  o/f«-girder  bridges  or  viaducts  the  area  for  the  wind- 

zuard  girder 
=  area  of  windward  face,  assumed  close,  between  rails 

and  top  of  train 
-f-  calculated  area  of  windward  surface  above  the  top 

of  the  train 
-f-  calctilated  area  of  windward  surface  below  the  rails. 
For  the  leeward  girder  or  girders  the  area 

=  calculated  area  of  surface  of  one  girder  above  the 
top  of  the  train  and  below  the  level  of  the  rails, 
the  pressure  being  28,  42,  or  56  lbs.  per  square 
foot,  according  as  this  area  <  f5,  >  \S  and  <  J5, 
or  >f  5,  where  5'  is  the  total  area  within  the  out- 
line of  the  girder.     The  assumed  factor  of  safety 
is  to  be  4. 
American   Specifications. — [ci)    The   lateral   bracing   iti   the 
plane  of  the  roadrvay  is  to  be  designed  so  as  to  bear  a  pressure 
of  30  lbs.  per  square  foot  upon  the  vertical  surface  of  one 
truss  and  upon  the  surface  of  a  train  averaging  12  sq.  ft., per 
lineal  foot,  i.e.,  360  lbs.  per  lineal  foot ;  this  latter  is  to  be  re- 
garded as  a  live  load.     The  lateral   'tracing  in  the  plane  of  the 
other  chord  is  to  be  designed  so  as  to  bear  a  pressure  of  50  lbs. 
per  square  foot  upon  tzoice  the  vertical  surface  of  one  truss. 
{b)  The  portal,  vertical,  and    horizontal  bracing  is  to  be 


654 


THEORY  OF  STRUCTURES. 


proportioned  for  a  pressure  of  30  lbs.  per  square  foot  upon 
twice  the  vertical  surface  of  one  truss  and  upon  the  surface  of 
a  train  averaging  10  sq.  ft.  per  lineal  foot,  i.e.,  300  lbs.  per 
lineal  foot,  the  latter  being  treated  as  a  live  load. 

(£•)  Live  load  in  plane    of   roadway  due    to  wind-pressure 
=  300  lbs.  per  lineal  foot. 
Fixed  load  in  plane  of  roadway  due  to  wind-pressure 

=  150  lbs.  per  lineal  foot. 
Fixed  load  in  plane  of  other  chord  due  to  wind-pressure 
=  150  lbs.  per  lineal  foot. 
Lateral  Bracing. — Consider    a    iruss-bridge   with   parallel 
chords  and  panels  of  length  p.     Let  A  be  the  area  of  the  ver- 
tical surface  of  one  truss. 

According  to  {a),  the  lateral  bracing  in  the  plane  of  the 
roadway  is  subjected  to  (i)  a  panel  live  load  of  360/  lbs.  and 
(2)  a  panel  fixed  load  of  30^  lbs.,  while  in  the  plane  of  the 
other  chord  it  is  subjected  to  a  panel  fixed  load  of 

50  X  2^1  =  \ooA  lbs. 


Thus,  if  the  figure  represent  the  bracing  in  the  plane  of  ihe 
roadway  of  a  ten-panel  truss,  and  if  the  wind  blow  upon  the 


I 


I 


8«5^^P     86j^p     segP     se^p     3«^P     Mj^p 
Fig.  416. 


side  AB,  the  maximum  horizontal  force  for  which  any  diagonal, 
e.g.-  CD,  is  to  be  designed  is 

=  4$A  lbs.  due  to  the  horizontal  force  of  30^!  lbs.  at 

each  panel  point 
-f-  756/  lbs.  due  to  the  horizontal  force  of  360/'  lbs.  at 
each  panel  point  between  C  and  B. 
The  dotted  lines  show  the  bracing  required  when  the  \yind 
blc"W3  on  the  opposite  side. 

It   is  sometimes    maintained   that    the  wind-forces  in  the  I 


aioKDs. 


655 


plane  of  the  upper  chords  of  a  through-bridge  or  the  lower 
chords  of  a  deck-bridge  arc  transmitted  to  the  floor-bracing 
through  the  posts.  This  can  hardly  be  correct  in  the  case  of 
long  posts,  as  they  do  not  possess  sufficient  stiffness.  It  lias, 
however,  been  pointed  out  by  Mr.  W.  B.  Dawson  that,  in 
through-bridges,  the  cumulative  effect  of  the  wind-pressure  at 
the  ends  of  the  bridge  might  produce  a  serious  bending  action 
ill  the  end  posts.  This  action  would  have  to  be  resisted  by 
additional  plating  on  the  end  posts  below  the  portals,  or  by  an 
increase  of  their  sectional  area. 

Under  wind-pressure  the  floor-beams  act  as  posts ;  hence, 
if  the  wind-bracing  is  attached  to  the  top  or  compression  flange 
of  a  floor-beam,  the  flange's  sectional  area  must  be  propor- 
tionately increased.  If  the  bracing  is  attaclicc^  to  the  lower 
or  tension  flange,  the  stresses  in  the  latter  will  be  diminished. 

22'  Chords. — The  wind-pressure  transmitted  through  the 
floor-bracing  increases  the  stresses  in  the  several  members,  or 
panel  lengths,  of  the  leeward  chord,  the  greatest  increments 
being  due  to  a  horizontal  force  of  (36q/>  -|-  30.i)  lbs.  at  each  of 
the  panel  points  in  AB.  The  corresponding  chord  stresses  in 
tile  ten-panel  truss-bridge  referred  to  above  are  : 

K,  =  0; 


C,  =  4K360/'  +  30/^)  tan  6  lbs. ; 
k=C  +  3K36o/+3O/0tan^ 

r.  =  C,  4-  2\{2>6op  +  zoA)  tan  ^ 
Y  =  C,  +  ik{l6op  +  2,oA)  tan  6 


8(360/'  +  30/i)  tan  B  lbs. ; 
101(360/  -f--  30/i)  tan  ii  lbs. ; 
12(360/'  +  30.i)  tan  ^  lbs. 


|50°— 6*  being  the  angle  between  a  diagonal  and  a  chord. 

Again,  the  wind-pressure  tends  to  capsize  a  train  and  throws 

l»n additional  pressure'of  P'^  lbs.  per  lineal  foot  upon  the  lee- 

pard  rail,  /"being  the  pressure  in  pounds  per  lineal  foot  on  the 
pin  surface,  y  the  vertical  distance  between  the  line  of  action 
t/'and  the  top  of  the  rails,  and  G  the  gauge  of  the  rails. 


% 


Mi 
ii 


m 


iilil 


li 


^5^  THEORY  OF  STRUCTURES. 

Thus,  the  total  pressure  on  leeward  rail 


and 


=  (-  -{- P  7^\  lbs.  per  lineal  foot, 

the  total  pressure  on  windward  rail 

(w  v\ 

2  ~  ^r)  ^^^-  P^*"  lineal  foot, 


w  being  the  weight  of  the  train  in  pounds  per  lineal  foot. 

Hence,  the  total  vertical  pressure  at  a  panel  point  of  the 
leeward  truss 


S-G 

2S 


S  being  the  distance  between  the  trusses. 

24.  Stringers. — Each  length  of  stringer  between  consecu- 
tive floor-beams  may  be  regarded  as  an  independent  girder 
resting  upon  supports  at  the  ends,  and  should  be  designed  to 
bear  with  safety  the  absolute  maximum  bending  moment  to 
which  it  may  be  subjected  by  the  live  load.  If  the  beams  are 
not  too  far  apart,  the  absolute  maximum  bending  moment  will 
be  at  the  centre  when  a  driver  is  at  that  point.  Again,  in  the 
case  of  the  Sault  Stc.  Marie  Bridge,  it  may  be  easily  shown 
that  the  maximum  bending  moment  is  produced  when  the  four 
pairs  of  drivers-  are  between  the  floor-beams. 

Let  J/  =  distance  of  first  driver  from  nearest  point  of  support. 

The  reaction  at  this  support 

=  J-|tr(824  -  Ay)  -  ^(206  -y). 

The  bending  moment  is  evidently  a  maximum  at  the  second 
or  third  driver,  and  at  the  second  driver 


=  J5^iL(2o6  —  j)(56  -{-y)  —  12000  X  56  ; 


MAXIMUM  ALLOWABLE   STRESS.  6$7 

at  the  third  driver 

=  ■5-§i(2o6  —  j/XioS  +/)  —  12000(52  +  108). 

In  the  first  case  it  is  an  absobite  maximum  when  y  =  75"  ; 


"     "    second  "     "    " 


J' =  49 


its  value  ill  each  case  being  2,i88,i66f  in. -lbs. 

Hence,  the  bending  moment  is  an  absolute  maximum  and 
tqual  to  2,i88,i66f  in.-lbs.,  at  two  points  distant  75  in.  from 
each  point  of  support. 

Also,  if  /,  is  the  moment  of  inertia  of  the  section  of  the 
stringer  at  these  points,  t,  the  distance  of  the  neutral  axis  from 
the  outside  skin,  and  /,  the  coefficient  of  strength,  then 


2  / 

-(2i88i66|)  =f,~  for  the  inner  stringer, 

3  ^1 


and 


-(2i88i66f)  =/,  -  for  the  outer  stringer. 
3  ^i 

The  continuity  of  the  stringers  adds  considerably  to  their 
:5t  length. 

25.  Maximum  Allowable  Stress. — Denoting  by  A  and  B, 
respectively,  the  numerically  greatest  and  least  stresses  to 
'A'hich  a  member  is  to  be  subjected,  the  following  rules  will  give 
results  which  are  in  accordance  with  the  best  practice : 

I.  Members  subjected  to  Tensile  Stresses  only. 

For  wrottght-iron,  maximum  stress  per  square  inch 

f         ^\  (  ^\ 

=  loooo  lbs.  =  8000'^  I  +  ^  j  lbs.  =:  ^3.81  4-  '-9  J  J  *^°"s- 

For  steel,  maxin\um  stress  per  square  inch 
=  12000  lbs.  =  10000(^1  +  -rj  lbs.  =  1 5.08  -\~  2.54-jj  tons. 


658 


THEORY  OF  STRUCTURES, 


II.  Members  subjected  both  to  Tensile  and  Compressive  Stresses. 

For  wronght-iron,  maximum  stress  per  square  inch 

=  8000^1  -  ^j  lbs.  =  ^3  81  —  1-9^)  tons. 

For  steel,  maximum  stress  per  square  inch 
=  10000(1  -  ~j  lbs.  =  [5.08  -  2.54 -^j  tons. 

III.  Members  subjected  to  Compressive  Stresses  only. 
Denote  the  ratio  of  the  length  (/)  to  the  least  radius  of 

gyration  {k)  by  r. 


The  maximum  stress  per  square  inch  = 


/ 


I  +  ar' 


lbs., 


/being  8000  lbs.  for  wrought-iron  and  to,ooo  lbs.  for  steel,  and 

-  being  40,000,  30,000,  or  20,000,  according  as  the  member  has 
a 

two  square  (fixed)  ends,  one  square  and  one  pin  end,  or  two 

pin  ends. 

Again,  the  maximum  stress  per  square  inch  for  steel  struts 

with  two  pin  ends  =  (loooo— 6or)(  i  -|-  -^j  lbs.; 
"  "  square  ends  =  (lOOOO— 4or)f  i  +  ^j  lbs.; 
«      "     pin  ends        =  (s  -  ^)(i  +  ;^)  tons; 

"      "     square  ends  =  f  5  —  ^](  1  +  -j-)  tons. 

In  the  last  two  expressions  r  <  40.     These  expressions  may 
be  also  employed  in  the  case  of  alternating  stresses,  but  the 

factor  must  then  be  changed  to  ( i  -j ^j. 


CAMBER. 


659 


26.  Camber. — Owing  to  the  play  at  the  joints,  a  girder  or 
truss  will  deflect  to  a  much  greater  extent  than  is  indicated  by 
tiicory,  and  the  material  will  receive  a  permanent  set,  which, 
however,  will  not  prove  detrimental  to  the  stability  of  the 
structure  unless  it  is  increased  by  subsequent  loads.  If  the 
chords  were  initially  made  straight,  they  would  curve  down- 
wards ;  and  although  it  does  not  necessarily  follow  that  the 
strength  of  the  truss  would  be  sensibly  impaired,  the  appear- 
ance would  not  be  pleasing. 

In  practice  it  is  often  specified  that  the  girder  or  truss  is  to 
nave  such  a  camber  or  upward  convexity  that  under  ordinary 
loads  the  grade  line  will  be  true  and  straight ;  or,  again,  that  a 
camber  shall  be  given  to  the  span  by  making  the  panel  lengths 
of  the  top  chord  greater  than  those  of  the  bottom  chord  by 
.125  in.  for  every  10  ft. 

The  lengths  of  the  web  members  in  a  cambered  truss  are 
not  the  same  as  if  the  chords  were  horizontal,  and  must  be  care- 
fully calculated  so  as  to  insure  that  the  several  parts  will  fit 
together. 

To  find  an  Approximate  Value  for  the  Camber,  etc. 

Let  d  be  the  depth  of  the  truss. 

Let  J, ,  J,  be  the  lengths  of  the  upper  and  lower  chords,  re- 
spectively. 

Let  /, ,  /,  be  thp  unit  stresses  in  upper  and  lower  chords, 
respectively. 

Let  </, ,  d^  be  the  distances  of  the  neutral  axis  from  the 
upper  and  lower  chords,  respectively. 

Let  R  be  the  radius  of  curvature  of  the  neutral  axis. 

Let  /  be  the  span  of  the  truss. 

Then 


R 


s.-l 


E 


and     ^  = 


R 


f^     .nnrnv 

-^,  approx., 


the  chords  being  assumed  to  be  circular  arcs. 

Hence,  the  excess  in  length  of  the  upper  over  the  lower 

chord 

/  d 


f: 


R 


■vk-.,-l'v-i.*t^J 


66o 


THEORY  OF  STRUCTURES. 


Let  4:,,  4r,  be  the  cambers  of  the  upper  and  lower  chord  . 
respectively;  /^ -f  </,  and  7?  —  d^,  are  the  radii  of  the  upiKi 
and  lower  chords,  respectively. 

By  similar  triangles, 

the  horizontal  distance  between  i        R-\-  d 
the  ends  of  the  upper  chord      j   ~       A'      ' 


the  horizontal  distance  between 
the  ends  of  the  lower  chord 


Hence, 


and 


(iR  +  d,y  /D  .    _.N 

\ "k — J  ~  ■**'  ■  ^'      '     '^'  ^PP''ox'"i^teIy, 

( o — '^J  —  -^a  •  ^C^  —  ^i),  approximately. 


27.   Rivet-connection   between    Flanges   and  Web  — 

The  web  is  generally  riveted  to  angle-irons  forming  part  of 
the  flanges. 

The  increment  of  the  flange  stress  transmitted  through 
the  web  from  point  to  point  tends  to  make  the  angle-irons  slide 
over  the  flange  surfaces. 

Denote  the  increment  hy  F,  and  let  ^  be  the  effective  depth 
of  the  girder  or  truss. 

Then,  if  5  be  the  shearing  force  at  any  point, 

Fh  =  the  increment  of  the  bending  moment  per  unit 
of  length 

=  (^)=5inthecaseofac.oseweb, 

and       Fh  =  the  increment  of  the  bending  moment 
2=  {^M)  s=  Sa  in  the  case  of  an  open  web ; 

a  being  the  distance  between  the  two  consecutive  apices  or 
panel  points  within  which  5  lies. 


EYE-BAHS  AX  J)  P/A/S. 


66 1 


Hence,  if  iV  be  the  nuiiibcr  of  rivets /tT  unit  of  length  for 
the  close  web,  or  the  number  between  the  two  consecutive 
apices  for  the  open  web, 

N f,  —  F  =  Y  for  the  close  web, 

4  fi 


and 


=  -j~  for  the  open  web, 


(/  being  the  diameter  of  ;i  rivet,  and  /,  the  safe  coefificient  of 
shearing  strength. 

28.  Eye-bars  and  Pins. — Eye-bars  connected  with  pins 
have  been  commonly  employed  in  the  construction  of  suspen- 
sion cables,  the  tension  chords  of  ordinary  trusses  and  canti- 
levers, and  the  diagonals  of  web  systems.  The  requisite  sec- 
tional area  is  obtained  by  placing  a  number  of  bars  side  by 
side  on  the  same  pin,  and,  if  necessary,  by  setting  two  or  more 
tiers  of  bars  one  above  another. 


j^ 


■^s^ 


Fig.  417. 


rBl 


1  I     ' 


n: 


tim 


t-rr 


T^3 L^I 


Fic.  418. 


43^ 

Fig.  419. 


The  figures  represent  groups  of  eye-bars  as  they  often 
occur  in  practice. 

If  two  sets  of  2«  bars  pull  upon  the  pin  in  opposite  direc- 
tions, as  in  Figs.  418  and  419,  the  bending  moment  on- the  pin 
will  be  nPp,  P  being  the  pull  upon  each  bar,  and /the  distance 
between  the  centre  lines  of  two  consecutive  bars. 


m 


IF 


lll'l 


I      1;     k-.    : 


.Ul 


662 


Hence, 


rilEOKV  OF  SI'h'UCrUKJiS, 


nVp      [r, 


f  bn'iij;  the  stress  in  tin-  material  of  the  pin  at  a  distance  c  fiDiu 
liu"  neutral  axis,  anil  /  thi-  moim-nt  nl  iiuitia. 

In  ji^ciiitiii,  the  bi-mlin};  ailion  iipm)  a  pin  loiiiuclin^f  ,i 
minihrr  of  vcrtieal,  hori/.oivlal,  ami  inelineil  b.irs  may  bo  »lc- 
termimHl  as  follows  : 

ConsitltM  Olio-half  of  the  pin  onlv. 

Let  /',  V'\^.  420,  be  the  losiiltant  stress  in  the  vertical  bars, 
It  is  nocossarily  otpial  in  maipiitmlo  but  opposite  in  (iiid. 
tion  to  the  vertical  component  of  the  resultant  of  the  stresses 


1 

-<  — 

1 

-1 

pl?^-, 

;i 

->H 


Fill.  490, 

in  the  inelineil  bars.  Let  r'  be  the  ih'stanee  between  the  liius 
of  action  of  these  two  resultants.  The  correspomliiif;  btiidiiti; 
action  upon  the  pin  is  that  due  to  a  couple  of  which  tin-  mo- 
ment is  /";•. 

Let  //  be  the  distance  between  the  lines  of  action  of  tiic 
equal  resultants  //  of  the  horizontal  stresses  upon  each  side  of 
the  pin.  The  correspondinj;  bending  action  upon  the  pin  is 
that  due  to  a  couple  of  which  the  moment  is  ////. 

Hence,  the  niaximum  bendin<;  action  is  that  due  to  a  couple 
of  which  the  moment  is  the  resultant  of  the  two  moments  Vv 
and  ////,  viz., 


Eyt'-bars. — In  Enp[land  it  has  been  the  practice  to  roll  bars 
having  enlarged  ends,  and  to  forge  the  eyes  under  hydraulic 


EYRHANS  AXn   I'/XS. 


U^i 


Fro.  431. 


prcssurr  with  siiil.ibl)'  .li.ipi d  dies.  In  America  botli  Ii;imiiit.*r- 
|cii|;ctl  .111(1  liydi.iiilir  forbid  cyo  |).irs  .iic  made,  tlic  lallfi  htin^j 
i.illcd  7tvA//('.v.v  rj'i-/'iifs,  Canfid  iiiallicnialit  al  and  fxpcriinrn- 
t.il  nivcsli^^.ilions  liavi-  l»rcn  »  an  icd  niit  tn  d(  tcrniiiM:  the  |»ii'|)r|- 
(linuMisiotis  of  III!  Iiid<-li(-.id  and  |)in,  hnt  owin;.;  to  lln  \iiy 
iiiniplcx  character  of  the  stie.sses  developed  in  t  iu'  nntal  .tionnd 
I'm   eye,  ,\u  accnratc  inathein.ilii  al  sohition  is  nnpossihle. 

I-et  //  he  tile  widlli  and  / 
tlir  tliiclvnesH  of  the  slutiik  of 
the  eye-har  represented  in  I'i};. 
,|.M.  I, el  .S'  he  the  widtii  of 
tin  metal  at  llie  sides  of  tlio 
lyr,  and  //  tlie  width  at 
end.  Let  I)  be  tlie  diameter 
(if  the  pin. 

The  proijortions  of  the  head 
nc  ^'overned  iiy  the  ^^eiieral  condition  that  each  and  every  part 
sliouM  i)e  at  least  as  stron;.;  as  the  shank. 

When  the  bar  is  snbjected  to  a  tensile  stress  the  pin  is 
ti;.;htly  embraced,  anil  failure  may  arise  from  any  one  of  tlic 
(ollowiii}^  causes  : 

(<■/)    The  pin  may  fie  shorn  throui^h. 

Hence,  if  the  pin  is  in  double  shear,  its  sectional  area  should 
be  at  least  one-half  i\\A\.  of  the  shank. 

It  may  happen  that  the  pin  is  bent,  but  that  fracture  is  pre- 
vented by  the  closinff  up  of  the  pieces  between  the  pin-head 
and  nut ;  the  efficiency, however,of  the  connection  is  destroyed, 
as  the  bars  are  no  lon}^'er  free  to  turn  on  the  pin. 

In  practice,  D  for  Hat  bars  varies  from  |c/ to  \(l,  but  usually 
lies  i)etvveen  J^/and  j|c/. 

The  diameter  of  the  pin  for  the  end  of  a  round  bar  is  gen- 
erally made  equal  to  i\  times  the  diameter  of  the  bar. 

The  pin  should  be  turned  so  as  tf)  fit  the  eye  accurately, 
but  the  best  practice  allows  a  difference  of  from  ^^  to  ^J^  of 
an  inch  in  the  diameters  of  the  pin  and  eye. 
{h)   The  link  may  tear  across  MN. 

On  account  of  the  perforation  of  the  head,  the  direct  pull 
on  the  shank  is  bent  out  of  the  straijjht  and  distributed  over 


Mil 
illlli 


'     t; 


m 


664 


T/mONy  OF  STNUCrVK/.S. 


the  sectiotis  X  There  is  no  rcasdii  for  the  assuinpliMn  ili,  t 
the  (h'sttihiilioii  is  uniform,  ami  it  is  oijviously  pidhalilr  iImi 
the  intensity  o{  stress  is  [greatest  in  the  metal  next  tiie  iuik. 
llcnce,  the  seetiomil  area  of  the  metal  aeross  JAV  must  he  ,it 
least  equal  to  that  of  the  shank,  ami  in  practice  is  alwa)' 
greater. 

5  usually  varies  from  .qi;//  to  .()2  5f/. 


The  sectional  area  throu^^h  the  sides  of  the  eye  in  the  he, id  ni 
a  round  bar  varies  from  li  limes  to  twice  that  of  the  bar. 

(( )  '/7/f  /•/«  tfutv  /'(•  torn  f/irottx/t  tlic  luiut 

rhiOt(tiuu7\\  the  sectional  area  of  the  metal  across  l\) 
should  be  cue  half  that  of  the  shank.  The  met.d  in  fnnit  dI 
the  pin.  h()\\ever,  may  be  likened  to  .i  uniformly  loailed  i.Mr(lcr 
with  l)oth  eiuls  fixed,  and  is  subjected  to  a  bemliin;  as  well  .is 
CO  a  shearinjT  action.  Hence,  the  itiiiiiiiiuni  value  of  //  h.i .  Iieni 
fixed  at  l(t  .uid  if  If  is  m;»de  e(|u.d  to  (/.  botli  kinds  ol  aelioii 
will  be  amply  jirovided  {ow 

{d)    The  lu (III II i^  surface  ntoy  i'e  if/SNj/reiei/f. 

If  such  be  the  case,  the  intensil)-  of  the  pressure  upon  the 
bearini;  surface  is  excessive,  the  eye  becomes  oval,  the  met.il  is 
upset,  and  a  fracture  takes  place.  Or  a^^ain,  as  tlu:  hole  elon- 
gates, the  metal  in  the  sections  vS"  n^'xt  the  hole  w  ill  be  diawii 
out,  aiul  a  crack  will  commence,  extending  outwards  until  li,u- 
ture  is  produced. 

In  practice,  adequate  bearing  surf.ice  m.i\-  be  obtained  by 
thickening  the  head  so  as  to  confine  the  maximum  intensity  of 
the  pressure  within  a  given  limit. 

(e)   The  head  may  he  torn  titrougli  the  shoulder  at  XV. 

Hence,  A'l'is  made  equal  to  d. 

The  radius  of  curvature  A'  of  the  shoulder  varies  from  ihl 

to  7.6</. 

d        2 
Note. — The  thickness  of  the  shank  .should  be   - ,  or     d  at 

4         7 

least. 

The  following  tabic  gives  the  eye-bar  proportions  common 
in  American  practice  : 


uvc  upon  the 
Uu-  nu't.il  !•> 

u:  hoU'  clou- 
iU  Uv  (h.wvii 
Is  \intil  li.u- 


STK/iL  EYEllARS. 


665 


Vahio 

Valiir 

Viilue  of  5. 

..(  ,/. 

..f  I). 

WrI.llrin 

lla'jimrrrd 

IIkir. 

M.UB. 

t     f)0 

.f.7 

'•5 

'■33 

1  .  nc  > 

.75 

I.S 

'•33 

1  .  (II ) 

1  .  1  If  1 

1  .=; 

t.'^.O 

T  .(Jll 

1  .l^ 

1  u 

«    50 

I  .no 

I    n 

•7 

1  .  '  1'  ) 

I  .  f,n 

>  H5 

1. 67 

1    III) 

'  •  Tn 

S.Od 

1.07 

1  .on 

a  ■  01 1 

■.■.2'^ 

'TS 



■ 



Alio,  111  WplillffiM  l);u«.  // ;     .S.    ill  li.iitimpicd  IjHrs,  II  ~d. 

29.  Steel  Eye-bars. — llydiiMilic-ffjr^'cd  steel  eye-bars  are 
iinu  heiiij.;  l.nj^cly  iiiiule.  Tin:  steel  has  an  (liliinate  teiiaeily  of 
tniin  60,000 to 68, OCK)  lbs.  i)cr  square  inch,  an  elastic  limit  of  not 
less  than  50  i)er  cent,  aiul  an  elonj^ation  of  from  17  to  20  per  cent 
ill  a  li'ii}.;th  e(|iial  to  A;/  times  the  least  transverse  dimension. 

The  riin-nix  Hridf^e  Co.  and  tlu'  Va\<^^ki  Moor  Iron  C<i.  jiive 
tin-  following  tables  of  steel  eye-bar  proportion.s  : 


I'licvnlx 

Hridgc  Co. 

Edge  Moor  Iron  Co. 

Minl- 

KxrpM  of 

Wi.lllK.f 

Illmiirlrr  of 
I'm  IkiIc. 

Dinin- 
riiT  iif 

Width 
..f 

MiHmcter 
of 

DiamclRr 

iniim 
TliKk- 

Sr(  tfiiti;il  Arra 
of  llriiil  al'iiiK 

b.ir  il. 

llcaU. 

bar  (/, 

I'in-liole. 

Heud. 

ncfiii 

/'/'ovrr  Sec- 

of Bar.  1 

tion  of  Har. 

3 

3,',v3|S 

7 

3 

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8 

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2  J 

54 

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54 

I 

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loi 

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f) 

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15 

5 

5ii 

124 

i 

7,1% 

7 

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IS 

f> 

5i 

134 

i 

31% 

7 

f'lS.  7i"„.  7 

lO 

() 

f'i 

144 

I 

21% 

7 

•  7 

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511 

■  54 

1! 

40% 

8 

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17 

7 

74 

17 

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8 

f>IJ.'>i8.  7i"„ 

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8 

7U.  Sil 

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18 

I 

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<) 

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20 

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81).  8J 

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22 

10 

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24 

In  l)oth  the   Plueiiix  and  Edge  Moor  bars  the  thickness  of  the  head  is  the 
same  as  that  of  the  Ixidy  of  the  bar,  or  does  not  exceed  it  by  more  than  ,'j  in. 


ittiii 


Zsmk'  ■ 


666 


THEORY  OF  STRUCTURES, 


30.  Rivets. — A  rivet  is  an  iron  or  steel  shank,  sHqlitlv 
tapered  at  one  end  (tlie  taW),  and  surmounted  at  the  otlici  by 
a  cup  ox  pan-shaped  head  {Vx^,  422).  It  is  used  to  join  steel  or 
iron  plates,  bars,  etc.  For  this  purpose  the  rivet  is  generally 
heated  to  a  cherry-red,  the  shank  or  spindle  is  passed  through 


^ 


Fio.  4ia. 


Fig.  433. 


Fig.  434, 


ZZii 


Fig.  425. 


Ku..  4j6. 


the  hole  prepared  for  it,  and  the  tail  is  made  into  a  button,  or 
point.  The  hollow  cup-tool  gives  to  the  point  a  nearly  hemi- 
spherical shape,  find  forms  what  is  called  a  snap-rivet  (Fig. 
423).  Snap-rivets,  partly  for  the  sake  of  appearance,  arc  com- 
monly used  in  girder-work,  but  tiiey  are  not  so  tight  as  conical. 
pointed  rivets  (j/<(/!^-rivets),  which  are  hammered  into  shape 
until  almost  cold  (Fig.  424). 

When  a  smooth  surface  is  required,  the  rivets  are  counter-sunk 
(Fig.  425).  The  counter-sinking  is  drilled  and  may  e.xteiul 
through  the  plate,  or  a  shoulder  may  be  left  at  the  inner  eilgc. 

Cold-riveting  is  adopted  for  the  small  rivets  in  boiler  work 
and  also  wherever  heating  is  impracticable,  but  tightly-drivcii 
turned  bolts  are  sometimes  substituted  for  the  rivets.  In  all 
such  cases  the  material  of  the  rivets  or  bolts  should  be  of  su- 
perior quality. 

Loose  rivets  are  easily  discovered  by  tapping,  and.  if  very 
loose,  should  be  at  once  replaced.  It  must  be  borne  in  niii)d, 
however,  that  expansions  and  contractions  of  a  complicated 
character  invariably  accompanj'  hot-riveting,  and  it  cainiot  be 
supposed  that  the  rivets  will  bj  perfectly  tight.  Intleed,  it  is 
doubtful  whether  a  rivet  has  any  hold  in  a  straight  drilled  hole, 
except  at  the  ends. 

Riveting  is  accomplished  either  by  hand  or  machine,  the  latter 
being  far  the  more  effective.  A  machine  will  s(|uee/e  a  rivet, 
at  almost  any  temperature,  into  a  most  irregular  hole,  but  the 
exigencies  of  practical  conditions  often  prevent  its  use,  except 
for  ordinary  work,  and  its  advantages  can  rarely  be  obtained 


Sl'KKArCTH  OF  PUNCHED  AND  DRILLED  PLATES.     667 

where  the)-  would  be  most  appreciated,  as,  e.tj.,  in  the  rivetitiij; 
up  of  connections. 

31.  Dimensions  of  Rivets. — Tlie  diameter  (<-/)  of  a  rivet 
in  ordinary  |jfirder-\vork  varies  from  \  in.  to  i  inch,  and  rarely 
exceeds  1^  in. 

The  thickness  {i)  of  a  plate  in  ordinary  tjirder-work  should 
never  be  less  than  \  in.,  and  a  thickness  of  f  in.,  or  even  -^^  in., 
is  preferable. 

Let  T  be  the  total  thickness  through  which  a  rivet  passes. 
According  to  Fairbairn, 

When  t  <\  in.,  d  should  be  about  2t. 
When  t  >  \  in.,  d  should  be  about  \^t. 
According  to  Unwin, 

When  /  varies  from  \  in.  to   i    in.   and   passes  through 
tivo  thicknesses  of  plate,  d  lies  between  \t  -\-  -^  and 

T      <; 

When  the  rivets  join  several  plates,  d  = (-   -. 

8       8 

According  to  I'^ench  practice, 

Diameter  of  head  =  i^d. 

Length  of  rivet  from  head  =  T -\-  l^d. 

According  to  Rankine, 

Length  of  rivet  from  head  —-T-\-  2^d. 

The  rise  of  the  heatl  =  %d. 

The  diameter  of  the  rivet-hole  is  made  larger  than  that  of 

the  shank  b\'  from  jV  to  -J  in.,  so  as  to  allow  for  the  expansion 

of  the  latter  when  hot. 

There  seems  to  be  no  objection  to  the  use  of  long  rivets, 

provided  they  are  properly  heated  and  secured. 

32.  Strength  of  Punched  and  Drilled  Plates. — Experi- 
ment shows  that  the  tenacity  of  iron  and  steel  plates  is  con- 
siderably diminished  by  punching.  This  deterioration  in 
tenacity  seems  to  be  due  to  a  molecular  change  in  a  narrow 
aniiulus  of  the  metal  around  the  hole.  The  removal  of  the 
annulus  largely  neutralizes  the  effect  of  the  punching,  and, 
hence,  the  holes  are  sometimes  punched  ^  in.  less  in  diameter 
than  the  rivets  and  are  subsequently  rimered  or  drilled  out  to 
the   full    size        The    original    stren;,';th    may    also    be    almost 


t:l 


III 


'.  ' -f  )\' 

'i  T'J7a&<1B^  H 

i 

'f 

11^ 

I 

'   k 

1 

!> 

V 

•       P 

k 

i: 

€68 


THEORY  OF  STRUCTURES. 


entirely  restored  by  annealing,  and,  generally,  in  steel  work, 
either  this  process  is  adopted  or  the  annulus  referred  to  above 
is  removed. 

Punching  does  not  sensibly  affect  the  strength  of  Landoie- 
Siemens  unannealcd  plates,  and  only  slightly  diminishes  the 
strength  of  thin  steel  plates,  but  causes  a  considerable  loss  of 
tenacity  in  thick  steel  plates  ;  the  loss,  however,  is  less  tlian 
for  iron  plates. 

The  harder  the  material  the  greater  is  the  loss  of  tenacity. 

Iron  seems  to  suffer  more  from  punching  when  the  lioles 
are  near  the  edge  than  when  removed  to  sonic  distance  from 
it,  while  mild  steel  suffers  less  when  the  hole  is  one  diameter 
from  the  edge  than  when  it  is  so  far  that  there  is  no  bulging  at 
the  edge. 

The  injury  caused  by  punching  may  be  avoided  b}'  drilling 
the  holes.  In  important  girder-work  and  whenever  great 
accuracy  of  workmanship  is  required,  a  uniform  pitch  nia\'  be 
insured  and  the  full  strength  of  the  metal  retained  by  the  use 
of  multiple  drills.  Drilling  is  a  necessity  for  first-class  work 
when  the  diameter  of  the  holes  is  less  than  the  thickness  of 
the  plate,  and  also  when  several  plates  are  piled.  It  is  impos- 
sible to  punch  plates,  bars,  angles,  etc.,  in  spite  of  all  ex- 
pedients, in  such  a  manner  that  the  holes  in  any  two  exactly 
correspond,  and  the  irregularity  becomes  intensified  in  a  pile, 
the  passage  of  the  rivet  often  being  completely  blocked.  A 
drift,  or  rimer,  is  then  driven  through  the  hole  by  main  force, 
cracking  and  bending  the  plates  in  its  passage,  and  separating 
them  one  from  another. 

The  holes  may  be  punched  for  ordinary  work,  and  in  plates 
of  which  the  thickness  is  less  than  the  diameter  of  the  rivets. 
Whenever  the  metal  is  of  an  inferior  quality,  the  holes  should 
be  drilled. 

33.  Riveted  Joints. — In  lap  joints  (Figs.  427  and  430)  the 
plates  overlap  and  are  riveted  together  by  one  or  more  rows 
of  rivets  which  are  said  to  be  in  single  shear,  as  each  rivet  has 
to  be  sheared  through  one  section  only. 

In  Jish  (or  butt)  joints  (Figs.  428  and  429)  the  rivets  are 
in  double  shear,  i.e.,  must  be  each  sheared  through  two  sections. 


RIVETED  JOINTS. 


669 


Thus  they  are   not  subjected  to  the  one-sided  pull  to  which 
rivets  in  single  shear  are  liable. 


IH 


Fig.  429. 


Fig.  430. 


\x\  fish  pints  the  ends  of  the  plates  meet,  and  the  plates 
are  riveted  to  a  single  cover  (Fig.  428),  or  to  two  covers  (Fig. 
429),  by  means  of  one  or  more  rows  of  rivets  on  each  side  of  the 
joint. 

A  fish  joint  is  properly  termed  a  butt  joint  when  the  plates 
are  in  compression.  The  plates  should  butt  evenly  against  one 
another,  although  they  seldom  do  so  in  practice.  Indeed,  the 
mere  process  of  riveting  draws  the  plates  slightly  apart,  leav- 
ing a  gap  which  is  often  concealed  by  caulkmg.  A  much 
better  method  is  to  fill  up  the  space  with  some  such  hard  sub- 
stance as  cast-zinc,  but  the  best  method,  if  the  work  will  allow 
of  the  increased  cost,  is  to  form  7i  jjwip  joint,  i.e.,  to  plane  the 
ejes  of  the  plates  carefully,  and  then  bring  them  into  close 
contact,  when  a  short  cover  with  one  or  two  rows  of  rivets  will 
suffice  to  hold  them  in  position. 

The  riveting  is  said  to  be  single,  donhh\  triple,  etc.,  according 
as  the  joint  is  secured  by  one,  two,  three,  or  more  rows  of  rivets. 


000 

000 

000 

000 

000 

000 

o°o 


o  o 

0% 


CHAIN 
Fig.  431. 


ZIGZAG 
Fig.  432. 


Double,  triple,  etc.,  riveting  may  be  chain  (Fig.  431)  or  zig- 
zag (Fig.  432).  In  the  former  case  the  rivets  form  straight 
lines  longitudinally  and  transversely,  while  in  the  latter  the 
rivets  in  each  row  divide  the  space  between  the  rivets  in  adja- 
cent rows.  Experiments  indicate  that  chain  is  somewhat 
stronger  than  zigzag  riveting. 


I  I 


it. 


670 


THEORY  OF  STRUCTURES. 


Figs.  433  to  435  show  forms  of  joint  usually  adopted  for 
bridge-work.  In  boiler-work  the  rivets  are  necessarily  very 
close  together,  and  if  the  strength  of  the  solid  plate  be  assumed 
to  be  100,  the  strength  of  a  single-riveted  joint  hardly  exceeds 
50,  while  double-riveting  will  only  increase  it  to  60  or  70.    Fair- 


Fig.  433. 


Fig.  434. 


Fig.  435. 


bairn  proposed  to  make  the  joint  and  unpunched  plate  equally 
strong  by  increasing  the  thickness  of  the  punched  portion  of 
the  plate,  but  this  is  somewhat  difficult  in  practice. 

The  stre.sses  developed  in  a  riveted  joint  are  of  a  most  com- 
plex character  and  can  hardly  be  subjected  to  exact  mathe- 
matical analysi.s. '  For  example,  the  distribution  of  stress  will 
be  necessarily  irregular  {a)  if  the  pull  upon  the  joint  is  one- 
sided ;  {b)  when  local  action  exists,  or  the  plates  stretch,  or  in- 
ternal straiiis  are  in  the  metal  before  punching;  {c)  if  there  is  a 
lack  of  symmetry  in  the  arrangement  of  the  rivets,  so  that  one 
rivet  is  more  severely  strained  than  another;  (d)  when  the 
workmanship  is  defective. 

The  joint  may  fail  in  any  one  of  the  following  ways  : 

(i)  The  rivets  may  shear. 

(2)  The  rivets  may  be  forced  into  and  crush  the  plate. 

(3)  The  rivets  may  be  torn  out  of  the  plate. 

(4)  The  plate  may  tear  in  a  direction  transverse  to  that  of 
the  stress. 

The  resistance  to  rupture  should  be  the  same  in  each  of  the 
four  cases,  and  always  as  great  as  possible. 

The  shearing  and  tensile  strengths  of  plate-iron  are  very 

nearly  equal.     Thus,  iron  with  a  tenacity  of  20  tons  per  square 

iarh  has  a  shearing  strength  of  18  to  20  tons  per  square  inch. 

!vv:t-iron  is  usually  somewhat  stronger  than  plate-iron. 

\gain,  the  shearing  strength  of  steel  per  square  inch  varies 


THEORETICAL  DEDUCTIONS. 


671 


from  .  30ut  24  tons  for  steel,  with  a  tenacity  of  about  30  tons, 
to  about  33  tons  for  steel,  with  a  tenacity  of  about  50  tons ;  an 
average  value  for  rivet-steel  with  a  tenacity  of  30  tons  being 
24  tons. 

Hence,  if  4  be  a  factor  of  safety,  the  working  coefficients 
become 

T^  ,  ^ .         (  5  tons  per  square  inch  in  shear,  and 

Tor  wroueht-iron  -(  -^  ^       ^ 

^  (  5     "       "        "         "      '•  tension. 


^            .  (6  tons  per  square  inch  in 

r  or  steel (  7Jr  "       "         "         "       *' 


in  shear,  and 
tension 


Allowance,  however,  must  be  made  for  irregularity  in  the  dis- 
tribution of  stress  and  for  defective  workmanship,  and  in 
riveting  wrought-iron  plates  together  it  is  a  common  practice 
to  make  the  aggregate  section  of  the  rivets  at  least  equal  to 
and  sometimes  20  per  cent  greater  than  the  net  section  of  the 
plate  through  the  rivet-holes. 

Hence,  the  working  coefficients  are  reduced  to 


and 


4  or  4^  tons  per  square  inch  for  wrought-iron, 

5  or  5i     "       "         "  "      "    steel. 


according  to  the  character  of  the  joint. 

There  is  very  little  reliable  information  respecting  the  in- 
dentation of  plates  by  rivets  and  bolts,  and  it  is  most  uncertain 
to  what  extent  the  tenacity  of  the  plates  is  affected  by  such 
indentation.  Further  experiments  are  required  to  show  the 
effect  of  the  crushing  pressure  upon  the  bearing  area  (i.e.,  tJie 
diameter  of  the  rivet  niultiplicd  by  the  thickness  of  the  plate], 
although  a  few  indicate  that  the  shearing  strength  of  the  rivet 
diminishes  after  the  intensity  of  the  bearing  pressure  exceeds 
a  certain  maximum  limit. 

34.  Theoretical  Deductions. 

Let  S  be  the  total  stress  at  a  riveted  joint ; 

/,,/,,/,,/,,  be  the  safe  tensile,  shearing,  compressive, 

and  bearing  unit  stresses,  respectively  ; 
t  be  the  thickness  of  a  plate,  and  w  its  width  , 


liil 

ill! 


I  .1 


-1  'it- 


672 


THEORY  OF  STRUCTURES. 


iVbe  the  total  number  of  rivets  on  one  side  of  a  joint; 

n  be  the  total  number  of  rivets  in  one  row ; 

p  be  the  pitch  of  the  rivets,  i.e.,  the  distance  centre  to 

centre ; 
d  be  the  diameter  of  the  rivets  ; 

X  be  the  distance  between  the  centre  line  of  the  nearest 
row  of  rivets  and  the  edge  of  the  plate. 
Value  of  x. — It  has  been  found  that  the  minimum  safe 
value  of  X  is  d,  and  this  in  most  cases  gives  a  sufificient  overlap 
{=  2x),  while  X  =  ^d  is  a  maximum  limit  which  amply  pro- 
vides for  the  bending  and  shearing  to  which  the  joint  may  be 
subjected.     Thus  the  overlap  will  vary  from  2d  to  ^d. 

X  may  be  supposed  to  consist  of  a  length  x,,  to  resist  the 
shearing  action,  and  a  length  x^  to  resist  the  bending  action. 
It  is  impossible  to  determine  theoretically  the  exact  value  of 
x^,  as  the  straining  at  the  joint  is  very  complex,  but  the  metal 
in  front  of  each  rivet  (the  rivets  at  the  ends  of  the  joint  ex- 
cepted) may  be  likened  to  a  uniformly  loaded  beam  of  length 

d 
d,  depth  x^ ,  and   breadth  /,  with  both  ends  Jixed.     Its 


(^>  -  t)  ' 


f  being  the 


/ 
moment  of  resistance  is   therefore  yrt 

6 

maximum  unit  stress  due  to  the  bending.     Also,  if  P  is  the 

load  upon  the  rivet,  the  mean  of  the  bending  moments  at  the 

P 

end  and  centre  is  -^d. 

o 

Hence,  approximately ^ 


i^^-ii  -  -=i^(-^)- 


It  will  be  assumed  that  the  shearing  strength  of  the  rivet 
is  equal  to  the  strength  of  a  beam  to  resist  cross-breaking. 
Single-riveted  lap  and  single-cover  joints  (Figs.  427  and  428). 


nd' 


/.  =  {p-d)tA  =  dtf,) 


.    .    (I) 


THEORE  TJCA  L  DED  UCTIONS. 


673 


8   / 


(2) 


^  ^  \  "      2/         4.  ■" 


.-.  ;«;,=  -  + 


I      /      d' 


f 


(3) 


As  already  pointed  out,  these  joints  are  weakened  by  the 
bending  action  developed,  and  possibly  also  by  the  concentra- 
tion of  the  stress  towards  the  inner  faces  of  the  plates. 

Single-riveted  double-cover  Joints  (Fig.  429). 


Ttd'  ^ 

^■4  ^' 


2^//=  =  2-—/,. 


{p-d)tf,=dtf,',    . 
nd' 


(4) 


•  •  *^  \ 


Id 


ndl 
4   t  ' 


(5) 


^/\* 


dV  _  I  na- 

~'2l    -2~A-^' 


^     ,     I 
'        2     '    4 


\/^r 


r 


(6) 


These  joints  are  much  stronger  than  joints  with  siiigle 
covers.  Also,  equation  (4)  shows  that  the  bearing  unit  stress 
in  a  double-cover  joint  is  twice  as  great  {theoretically)  as  in  a 
single-cover  joint  (eq.  i),  so  that  rivets  of  a  larger  diameter 
may  be  employed  in  the  latter  than  is  possible  in  the  former, 

for  corresponding  values  of  -. 


m 


t^: ; 


674  THEORY  OF  STRUCTURES. 

Chain-riveted  joints  (Fig.  431). 

fiw-nd)t  =  S  =  f,Ndt\ 

_  y.r'^d 

S  =  N  — /^  when  there  is  one  cover  only ;    . 
4 

nd' 
S  =  N /,  when  there  are  two  covers.    .     .    . 


This  class  of  joint  is  employed  for  the  flanges  of  bridge- 
girders,  the  plates  being  piled  as  in  Figs.  436,  437,  438,  and  ;/ 
being  usually  3,  4,  or  5. 

In  Fig.  437  the  plates  are  grouped  so  as  to  break  joint,  and 
opinions  differ  as  to  whether  this  arrangement  is  superior  to 
the  full  butt  shown  in  Fig.  438.     The  advantages  of  the  latter 

/.^/~^^-^  r^  r^  r\  r>  /-\ 


I.e.,  a 

(7) 

tucen 
Le 

(8) 

13,  •  ■ 
Til 

dently 

(9) 

Let 
sile  strt 

2  2,3i 

^ 


X 


3 


Fig.  436. 

I Ci Ci — Ci Q lO Ci Q Q_ 


X 


ru 


X 


Fig.  437. 


^3- 


"C 


"O  \J     KJ     C7  ^^~~0 — CT" 
Fig.  438. 


are  that  the  plates  may  be  cut  in  uniform  lengths,  and  the 
flanges  built  up  with  a  degree  of  accuracy  which  cannot  be 
otherwise  attained,  while  the  short  and  awkward  pieces  accom- 
panying broken  joints  are  dispen.sed  with. 

A  good  practical  rule,  and  one  saving  much  labor  and  ex- 
pense, is  to  make  the  lengths  of  the  plates,  bars,  etc.,  multiples 
of  the  pitch,  and  to  design  the  covers,  connections,  etc.,  so  as 
to  interfere  with  the  pitch  as  little  as  possible. 

The  distance  between  two  consecutive  joints  of  a  group 
(Fig.  437)  is  generally  made  equal  to  t^vice  the  pitch. 

An  excellent  plan  for  lap  and  single-cover  joints  is  to 
arrange  the  rivets  as  shown  in  Figs.  431  to  435. 

The  strength  of  the  plate  at  the  joint  is  only  weakened  by 
one  rivet-hole,  for  the  plate  cannot  tear  at  its  weakest  section, 


Assuir 
Hence 

above  rek 

A,  the  ass 
35.  Co 

must  not  1 
a  si//£-le  CO 
if  there  an 

When 
happens  th 
the  greatei 


COVERS.  675 

i.e.,  along  the  central  row  of  rivets  {an),  until  the  rivets  be- 
tween it  and  the  edge  are  shorn  in  two. 


Let  there  be  m  rows  of  rivets,  i  1,22, 


33,...  (Fig.  439). 

The   total    number   of    rivets 
dcntly 


IS   evi- 


VI' 


Let /,,^,,^3,^, 


be  the  unit  ten- 


sile stresses  in  the  plate  along  the  lines  i  i, 


2  2,  3  3, 


respectively.     Then 


nd' 


3  2  I 


3  2 

Fio.  439. 


S  =  {w-  d)tf^  =  —-  ;«y; ,         for  the  line  i  i ; 

4 


I  i 


11 


4 


2  2  ; 


4 


3  3i 


ill! 


Ttd^ 


=  (w-4^)/^,  =  — -( 


m 


6y, 


S  =  (w  —  d)tf,  =  {w  —  2d) 


m 


m  —  I 


;^?,  = 


44; 


t\  > 


are  each  less  than 


Assume  that/,  =  q^ .     Then  w  =.  {in'  +  i)^. 

Hence,  by  substituting  this  value  of  iv  in  the  first  of  the 

above  relations,  -  =  —  -9.    Since  g. ,  q, 

t  II      /,  33.^4 

/, ,  the  assumption  is  justifiable. 

35.  Covers. — In  tension  joints  the  strength  of  the  covers 
must  not  be  less  than  that  of  the  plates  to  be  united.  Hence, 
a  single  cover  should  be  at  least  as  thick  as  a  single  plate  ;  and 
if  there  are  two  covers,  each  should  be  at  least  half  as  thick. 

When  two  covers  are  used  in  a  tension  pile  it  often 
happens  that  a  joint  occurs  in  the  top  or  bottom  plate,  so  that 
the  greater  portion  of  the  stress  in  that  plate  may  have  to  be 


p! 


676 


THEORY  OF  STRUCTURES. 


borne  by  the  nearest  cover.     It  is,  therefore,  considered  advis- 
able to  make  its  thickness  five-eighths  that  of  the  plate. 

The  number  of  the  joints  should  be  reduced  to  a  minimum, 
as  the  introduction  of  covers  adds  a  large  percentage  to  the 
dead  weight  of  the  pile. 

Covers  might  be  whollj'  dispensed  with  in  pcrfcct-jnmp 
joints,  and  a  great  econoni)'^  of  material  effected,  if  the  dif- 
ficulty  of  forming  such  joints  and  the  increased  cost  did  not 
render  them  impracticable.  Hence,  it  may  be  said  that  covers 
are  required  for  all  compressicn  joints,  and  that  they  must  be  as 
strong  as  the  plates ;  for,  unless  the  plates  butt  closely,  the 
whole  of  the  thrust  will  be  transmitted  through  the  covers. 
In  some  of  the  best  examples  of  bridge  construction  the  ten- 
sion and  compression  joints  are  identical. 

36.  Efficiency  of  Riveted  Joints.* — The  efficiency  of  a 
riveted  joint  is  the  ratio  of  the  maximum  stress  which  can  be 
transmitted  to  the  plates  through  the  joint  to  the  strength  of 
the  solid  plates. 

Denote  this  maximum  efficiency  by  r}. 
Let  /  be  the  pitch  of  the  rivets  ; 

d      "       diameter  of  the,  rivets  ; 
/       "       thickness  of  the  plates  ; 
ft     "       tenacity  of  the  solid  plate ; 
mft  "  "         "     "  riveted  plate ; 

f,     "       shearing  strength  of  the  rivets  ; 
N     "       number  of  rivets  in  a  pitch  length  ; 
e      "      ratio  of  the    strength  of  a  rivet  in  double 
shear  to  its  strength  in  single  shear. 


Then 


I/,  =  efficiency  as  regards  the  plates  =  — J. — - 

^^P_=J),      (I) 


»;,  =  efficiency  as  regards  the  rivets  = 


eN-d'f. 
4 

Ptft      ' 


*  From  an  article  by  Professor  Nicolson  in  the  Engineer,  Oct.  9,  ifi 


(2) 


EFFICIENCY  OF  RIVETED  JOINTS. 


677 


The  eflficiency  of  the  joint  is,  of  course,  the  sr,aller  of  these 
two  values ;  and  the  joint  is  one  of  maximum  efficiency  when 
»/,  =  1;,  =  V  ;  that  is,  when 


nt 


p  —  d__  4_ ^ 


or 


.71 


{p-d)tmf,  =  cN-dV. 
4 


(3) 


In  this  expression  the  quantities  m  ft,  N,  and  c  are  con- 
stants for  any  given  joint,  being  of  necessity  known,  or  having 
been  fixed  beforehand  ;  and  the  equation  thus  expresses  one 
condition  governing  the  relations  of  the  three  variables  p,  d, 
and  t  to  each  other.  It  is  obvious,  however,  that,  in  order  to 
determine  the  values  of  any  two  of  these  variables  in  terms  of 
the  third,  another  relation  between  them  must  be  postulated. 
In  short,  in  designing  a  joint,  the  value  of  one  of  the  three 

ratios  ~,  -,  and  -  must  be  fixed. 
at  t 

■     P 
Case  I.    Suppose  that   the   ratio  -,  has  a  certain  value. 

This  is  very  frequently  the  quantity  predetermined ;  but  it  is 
most  usually  done  by  fixing  the  value  of  v>  V  very  obviously 

P  r  I  A 

volving  -,',  m  fact  Tf=m\i  —  -y. 


in 


d'  '      ''\'       pi 

Equation  (3)  may  be  written 

p  ^  eN'^d' A  j^  d, 
^  4   t  mfi   ' 


or 


P  =  d[eN~%^^-\-'^ (4) 


d 


If  the  ratio  -  be  denoted  by  k,  then 


mu 


d  4    '«/* 


(5) 


678  THEORY  OF  STRUCTURES. 

_,         ;  '        m{p  —■  d)  '       ' 

But  Since  7  =         ^ -, 

P 

P  _      m 

1~  m-r, (^) 

Therefore,  substituting  in  (5), 

and,  ultimately, 

'^~  eNTt   f/m-v ^^^ 

The  process  of  designing  a  joint  of  maximum  efficiency  for 
a  boiler  of  given   diameter  and   pressure  of  steam,  when   /; 

for  the  ratio  ^j  is  fixed,  is  then  as  follows :  Settle  the  number 

of  rivets  per  pitch  (i.e.,  N) ;  the  value  to  be  allowed  for  e  (de- 
pending on  the  nature  of  the  shearing  stress  on  the  rivets^ ; 
and  the  values  of  m,fif  and  /,.  Then  k  is  known  from  equa- 
tion (8). 

But  /  may  be  found  from  the  relation, 

pressure  X  diameter  =  rf  X  2t/t , 
or 

pressure  X  diameter 

/  = ^^ •    •    •    ...    (9^ 

Hence,  since  ^  =  —  is  known,  d  may  be    found ;   and  since 

-.  = is  known,  p  is  also  fixed.  ":      . 

d      m  —  1?  -^  :.■■■■■■■'. 

P 
Case  II.    When  -,  the  ratio  of  rivet  pitch  to  plate  thick- 

ness,  is  given,  equation  (5)  must  be  otherwise  manipulated. 


Mu 
have 


Putting 


For  bre 
and  solv 


Ther 
A,  T, 

their  valu 
term  beii: 
Now, 


and  since 
plate  (/)  : 
values  of 
the  knowi 
This  n 
rational  ol 
will  remai 
of  rivets  t( 
the  relativ 

Case 

must  first 


ii'li 


ff-  ( 


EFFICIENCY   OF  RIVETED  JOINTS. 


6;9 


Multiplying  it  by  -,  and  substituting  for  d  its  value  kt,  we 
P 


have 


.  =  £^'i*.A+i. 


4    P     I'ifi       P 
Putting  this  in  the  form  of  a  quadratic  equation  in  k, 


(10) 


eNn  f  cNn  f,    t 


.     .     (II) 


For  brevity,  substituting  A  for  —ry- ,  T  for  -j- ,  and  R  for  -^ 
and  solving  the  quadratic, 

AT      I 


k=^-—±-\fAT-^AA  TR.  .     .    .    (12) 


^f'lii 


The  method  of  designing  the  joint  is,  then,  as  follows : 

A,  T,  and  R  being  known,  k  may  be  found  by  substituting 

their  values  in  equation  (12),  the  positive  sign  of  the  second 

term  being  taken. 
Now, 

ff  =  m{l  -j)  =  m{i-^)  =  ^(i  -|)  ; 

and    since  both  k  and  R  are  now  known,   the    thickness   of 

plate  {()  may  be  found,  as  in  Case  I,  by  equation  (9),     The 

values  of  the  diameter  and  pitch  of  rivets  follow  at  once  from 

the  known  values  of  k  and  R. 

This  method  of  designing  a  joint  appears  to  be  the  most 

rational  of  the  three.     For  the  greatest  pitch  for  which  a  joint 

will  remain  steam-tight  depends  mainly  on  the  relation  of  pitch 

of  rivets  to  thickness  of  plates ;  although  it  is  also  affected  by 

the  relative  size  of  rivets  and  of  rivet-heads. 

d 
Case   III.    If  — ,  or  k,  be  predetermined,  the  value  of  7 

must  first  be  obtained,  in  order  that  the  plate  thickness  may 


ill.   I 


68o 


THEORY  OF  STRUCTURES. 


P  ~  d 
be  found  by  means  of  equation  (9).     Now,  r}  =  m- may 

be  put  into  the  form 


/  = 


tnd 


and  if  this  value  is  substituted  for/  in  equation  (4), 


md 


m 


—  V        \    A 


V 


From  this  is  finally  deduced 


rf  =.  m 


Vlft 


eNnkf 


)d 


eNnkf,  +  4w// 


(13) 


The  plate  thickness  may  now  be  found  by  equation  (9); 
the   diameter    of    rivet    from   d  =  kt,    and    the    pitch   from 

p  = .     In  the  above  investigations  no  account  has  been 

^       m—  rf  ° 

taken  of  the  effect  of  the  bearing  pressure  on  the  rivets  or 

plate. 

li  fc  be  the  allowable  bearing  pressure  per  projected  square 

inch  of  rivet  surface,  the  following  relation  must  obtain  : 


{p-d)tmf,^Ndtf, (14) 


This  may  be  written 


/  = 


{p  —  d)mft 

Nd  ^  ' 


(IS) 


Then  if/,,  be  estimated  by  this  equation,  and  if  it  should  be 
greater  than  43  tons  per  square  inch  in  a  lap  joint,  or  45  to  50 
tons  in  a  butt  joint,  such  joint  will  fail  by  the  rivets  shearing 
before  the  full  strength  of  the  plate  is  exerted,  as  Kennedy's 
experiments  show  that  with  these  values  of  f  the  rivets  do 
not  attain  their  natural  ultimate  shearing  strength  (viz.,/,),  but 
fail  at  shearing  stresses  much  beL  .v  this. 


Again 

prelimina 
using  the 


deduced  fi 


{Unwin  suj 

In  desij 
any  value 
should  be  r 


Note.- 
have  been  f 
bridges  in  q 


EFFICIENCY  OF  RIVETED  JOINTS. 


d 


68 1 


Again,  the  maximum  allowable  ratio  of  -  (i.e.,  K)  as  the 

preliminary  datum  for  the  design  of  a  joint,  may  be  fixed  by 
using  the  expression 

deduced  from  the  obvious  relati'^n — similar  to  (14)^ 


(16) 


eN-d'f,  =  Ndtf,. 
4 

(Unwin  suggests  the  relation  d=:.  ^  V^.) 

In  designing  the  joint  by  any  of  the  methods  given  above, 
any  value  obtained  for  k  greater  than  that  supplied  by  (16) 
should  be  rejected. 


* 


m 


Itii:. 


Note, — The  following  Tables  of  the  Weights  of  Bridges 
have  been  prepared  from  data  supplied  by  the  engineers  of  the 
bridges  in  question. 


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THEORY  OF  STRUCTURES. 


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TABLE  OF  ACTUAL    WEIGHTS  OF  MODERN  BRIDGES.    687 


si  i 


i|t' 


|l 


\i 


688  THEORY  OF  STRUCTURES. 

TABLE   OF   LOADS   FOR   HIGHWAY   BRIDGES. 


Span  in  Feet. 


loo  and  under 
loo  to  200 
200  to  300 
300  to  400 
above  400 


City  and  Suburban 
Bridges  liable  to 
Heavy  Traffic. 


Bridi^es  in  Manu- 
facturing Districts. 
Ballasted  Koad. 


Bridges  in  Country 

Districts, 
Unballasted  Koads. 


100  lbs.  per  sq.  ft. 
80  ••   " 
70  "   " 
60  "   " 
50 


go  lbs.  per  sq, 
60  "   " 
50  "   " 
50  "   " 
50  "   " 


ft. 


70  lbs.  per  sq.  ft. 
60  "   " 
50  "   " 
45  "  " 
45  "  " 


^ 


EXAMPLES. 


689 


1^ 


EXAMPLES. 

I.  A  bridge  of  A^ equal  spans  crosses  a  span  of  Z  ft.;  the  weights  in 
tons  per  lineal  foot  of  the  main  girders  of  the  platform,  permanent  way, 
etc.,  and  of  the  live  load,  are  wi ,  Wa ,  71/, ,  respectively.     Show  that 


LA 


w,  = 


N-  LB' 


where  A  =  Wt^pk  +  ^)  +  iv%{pk  +  q)    and    B  =pk  ■\-  r, 

k  being  the  ratio  of  span  to  depth,  and  p,  q,  r  numerical  coefficients. 
Hence  also  determine  the  limiting  span  of  a  girder. 

If  X  is  the  cost  of  a  pier  and  if  Y  is  the  cost  per  ton  of  the  super- 
structure, find  the  va'.ue  of  iV  which  will  make  the  total  cost  per  lineal 
foot  a  maximum,  and  prove  that  this  is  approximately  the  case  when  the 
spans  are  so  arranged  that  the  cost  of  one  span  of  the  bridge  structure 
is  equal  to  the  cost  of  a  pier. 

Ans.  A  span  < 77-7-^    Cost  is  a  minimum  when  A''  —  LB  =  Z4/__, 


pk  -^  q. 


and  the  minimum  cost  of  the  span 


-(-f) 


X,  approx. 


2.  A  car  of  weight  W  for  a  gauge  of  4  ft.  81  in.  is  33  ft.  long,  6  ft.  deep, 
and  its  bottom  is  2  ft.  6  in.  above  the  rails.  Find  the  additional  weight 
thrown  upon  the  leeward  rail  when  the  wind  blows  upon  a  side  of  the 
car  with  a  pressure  of  20  lbs.  per  square  foot.  Also  find  the  minimum 
pressure  that  will  blow  the  car  over.  Ans.  4625.84  lbs. ;  .428  W. 

3.  A  lattice-girder  200  ft.  long  and  20  ft.  deep,  with  two  systems  of 
ri^'ht-angled  triangles,  carries  a  dead  load  of  800  lbs.  per  lineal  foot. 
Determine  the  greatest  stresses  in  the  diagonals  and  chords  of  the  fourth 
bay  from  one  end  when  a  live  load  of  1200  lbs.  per  lineal  foot  passes  over 
the  girder. 

Ans.  \i  riveted:  Diagonal  stress  =  37,2004/2  lbs. ; 

Chord  stress  =  450,000  lbs.  _ 

\{ pin-connected :  Diagonal  stress  =  44,800 1^2  and  29,600  V2  lbs.; 
Chord  stress  =:  460,000  lbs. 


i 


1:!^ 


I  t  1 


690 


THEORY  OF  STRUCTURES. 


4.  A  latticc-girdcr  80  ft.  long  and  8  ft.  deep  carries  a  uniformly  dis- 
tributed load  of  144,000  lbs.  Find  the  flange  inch-stresses  at  the  centre, 
the  sectional  area  of  the  top  flange  being  56i  sq.  in.  gross,  and  of  ilie 
bottom  flange  45  sq.  in.  net. 

What  should  be  the  camber  of  the  girder,  and  what  extra  length 
should  be  given  to  the  top  flange,  so  that  the  bottom  flange  of  the  loaded 
girder  may  be  truly  horizontal  ?     {E  =.  29,000,000  lbs.) 

Ans,  3185.8  lbs.  ;  4000  lbs. 

x\  =  .29735 ;  -t'  =  •  2987  ;  J.  -  5,  =  jYM- 

5.  A  lattice-girder  80  ft.  long  and  10  ft.  deep,  with  four  systems  of 
right-angled  triangles,  carries  a  dead  load  of  1000  lbs.  per  lineal  foot. 
Determine  the  greatest  stresses  in  the  diagonals  met  by  a  verticul  plane 
in  the  seventh  bay  from  one  end  when  a  live  load  of  2500  lbs,  per  lineal 
foot  passes  over  the  girder.  Design  the  flanges,  which  are  to  consist  of 
plates  riveted  together. 

The  lattice-bars  are  riveted  to  angle-irons.  Find  the  number  of  |-in, 
rivets  required  to  connect  the  angle-irons  with  the  flanges  in  the  first 
bay,  10,000  lbs.  per  square  inch  being  the  safe  shearing  strength  of  the 
rivets. 

Ans.  \i  riveted:  Diagonal  stress  =  io,664-,ij|/2  lbs. 

\{ pin-connected :       "  =  9062^1/2;  6250V2;  15,4684 V"2; 

ii,87sV2  lbs. 
22  rivets  (2i-i\). 

6.  The  bracing  of  a  lattice-girder  consists  of  a  single  system  of  tri- 
angles in  which  one  of  the  sides  is  a  strut  and  the  other  a  tie  inclined  to 
the  horizontal  at  angles  of  a  and  fi  respectively ;  in  order  to  give  the 
strut  sufficient  rigidity  its  section  is  made  i  times  that  indicated  by 
theory,  the  coefficient  I'  being  >  unity.  Show  that  the  amount  of  ma- 
terial in  the  struts  and  ties  is  a  minimum  when 

tan  a  =  i  tan  /3. 

7.  A  lattice-girder  of  40  ft.  span,  5  ft.  depth,  and  with  horizontal 
chords  has  a  web  composed  of  two  systems  of  right-angled  triangles  and 
is  designed  to  support  a  dead  and  a  live  load,  each  of  i  ton  per  lineal 
foot.  Determine  the  maximum  stresses  in  the  members  of  the  third 
bay  from  one  end  met  by  a  vertical  plane. 

Ans.  If  riveted:  Diagonal  stress  =  ^^2  tons ; 

Chord  stress  =  27  tons. 
If  pin-connected :  Diagonal  stress  =  \\^2  and  fjV2  tons ; 
Chord  stress  =  26  tons. 

8.  A  lattice-truss  of  100  ft.  span  and  10  ft.  depth  has  a  web  composed 
of  four  systems  of  right-angled  triangles.     The  maximum  stress  in  the 


EXAMPLES. 


691 


II II 


(iiagonal  joining  the  sixth  apex  in  the  upper  chord  to  the  fourth  apex  in 
The  lower  is  16  tons.  F"ind  the  dead  load,  the  live  load  bring  i  ton  per 
lineal  foot,  assuming  the  truss  to  be  («)  riveted,  (b)  pin-connected. 

Ans.   (a)  .554  ton  ;  {6)  1.062  tons. 

9.  A  lattice-girder  of  40  ft.  span  has  a  web  composed  of  t\/o  systems 
of  triangles  (base  =  10  ft.)  and  is  designed  to  carry  a  live  load  of  i6oo 
.us.  per  lineal  foot  and  a  dead  load  of  1200  lbs.  y^er  lineal  foot.  Defin- 
ing the  stress-length  of  a  member  to  bf  the  product  of  its  length  into 
tiie  stress  to  which  it  is  subjected  find  the  depth  of  the  truss  so  that  its 
A'/.// stress-length  may  be  a  minimum.  Ans.   10.19  ^t. 

10.  Determine  the  maximum  stresses  in  the  members  of  a  lattice- 
truss  of  40  ft.  span  a;id  4  ft.  depth,  with  two  systems  of  triangles  (base 
=  8  ft.),  (a)  when  riveted  together;  (d)  when  pin-connected.  Dead  load 
—  i  ton  per  lineal  foot,  live  load  =  i  ton  per  lineal  foot. 

Ans.  Bays —  ist; 
(<i)  Diags.        e^vT; 
Tens.chd.      6J ; 
Comp.  chd. 

(/.)  Diags.    \^^];J 
(7-5V2 

Tens.chd.      6; 
Comp.  chd. 

11.  The  platform  of  a  single-track  bridge  is  supported  upon  the  top 
chords  of  two  Warren  girders  ;  each  girder  is  100  ft.  long,  and  its  brac- 
ing is  formed  of  ten  equilateral  triangles  (base  10  ft.);  the  dead  weight 
of  the  bridge  is  900  lbs.  per  lineal  foot. ;  the  greatest  total  stress  in  the 
seventh  sloping  member  from  one  end  when  a  train  crosses  the  bridge 
is  41,394.8  lbs.  Determine  the  weight  of  the  live  load  per  lineal  foot. 
Prepare  a  table  showing  the  greatest  stress  in  each  bar  and  bay  when  a 
single  load  of  1 5,000  lbs.  crosses  the  girder. 

Ans.        ti^ft  <>    h    r 

^\;^V(\AAA/\/\/\A/         277if  lbs.  per  lin.  ft 
«i    <^    <:i    M 

Fig.  440. 

Stresses  in  diagonals  :      t/?  =  rfj  =  9  V3  ;    </s  =  c'l  =  8  4/3 ; 

^5  r=  ^9  =  7  1/3  ;    ^,  =  </„  =  6  4/3 ; 

</»  =</io=  5  1/3  tons. 
Stresses  in  compression  :  fi  =  9  i/3  ;  ca  =  16  V'3  ;  r,  =21  //J; 

f 4  =  24  4/3  tons. 

Stresses  in  torsion :  /,  =  4^  4/3  ;  /j  =  1 2  4/3";  /« =  1 7i  4/3 ; 

/4  =  21  13  ;  /6=22i  y'3  tons. 


2d; 

3fi; 

4th; 

5th. 

5-53^^; 

4.054/2 

t 

2.854/2 

• 

If  4/2" tons ; 

i8f; 

274; 

Same. 

33l: 

364       " 

4-7^2  f 

j  3-4V2'& 
'1  4.7Vi" 

H 

3-44/2'& 
2.31/3 

M' 

•2^^*itons; 

18; 

27; 
Same. 

33; 

36 

illl 


ii 


692 


THEORY  OF  STRUCTURES. 


12.  A  Warren  girder  with  its  bracing  formed  of  nine  equilateral  tri- 
angles (base  =  10  ft.)  is  90  ft.  long,  and  its  dead  weight  is  500  lbs.  per 
lineal  foot.  Determine  the  maximum  stresses  in  each  member  when 
a  live  load  of  1350  lbs.  per  lineal  foot,  preceded  by  a  concentrated 
load  of  18,000  lbs.,  crosses  tlie  girder,  assuming  that  every  joint  is  loaded. 
The  diagonals  and  verticals  are  riveted  to  angle-irons  forming  part  of 
the  flanges. 

How  many  f-in.  rivets  are  required  for  the  connection  of  the  several 
members  meeting  at  the  third  apex  in  the  upper  chord  ?  (23,  6,  and  13.) 
How  many  are  required  in  the  first  bay  of  each  chord  to  prevent  longi- 
tudinal slip?     (15  in  tension  chord  and  18  in  compression  chord.) 


Ans, 


Pig.  441. 

Tension  chord  stresses : 

.       i        87125  .                230625 

V3                  Vi 

.  _  .  _  333125.  .    ,    394625 

^/^                 V3 

^3 

^  .        ,       ,  164000  287000  •;6qooo 

Compression  chord  stresses :  c\  =  — ._ -;  ct  —  — ;=-  ;  Ci-^"^——  ; 

V3  \l  V3 

410000  ,. 
c^  =  —-1=-  lbs. 

V3 

_  .     ,     .  .       178500     ,     159500    ,     141250 

Stresses  in  slopmg  members:  «i=  — p^  ;  «a= — ;=^  ;  «3= — -p—  ; 

1^3       i^3       4/3 

123750     107000    91000 
V3  V3  V3 


./   7^750.  . 
V3 


70250   .  47500 

V3  V3 


34500    22250    10750 

S/3  ^3  V3 

lbs.  The  stresses  ^10,  dn  ,  dii,  are 
max.  stresses  of  an  opposite  kind 
to  those  due  to  dead  load. 

Verticals :   Max.  load  on  each  vertical  =  20, 500  lbs. 
13.  If  a  force  of  5000  lbs.  strike  the  bottom  chord  of  the  girder  in  the 
preceding  question  at  20  ft.  from  one  end  and  in  a  direction  inclined  at 
30°  to  the  horizontal,  determine  its  effect  upon  the  several  members. 


EXAMPLES. 


693 


394625 

V3'    • 


141250 

'~    V3    ' 
91000 

_ 47500 

10750 

"  4/3' 

,  rfn,  are 
■jsite  kind 


14.  A  Warren  girder  for  a  single-track  railway  bridge  consists  of 
eiglit  equilateral  triangles  and  has  to  cross  a  span  of  96  ft.  ;  the  platform 
is  on  the  bottom  chord  ;  the  loads  per  lineal  foot  for  which  the  truss  is  to 
be  designed  are  2250  lbs.  due  to  engine.  1 500  lbs.  due  to  traiiv  and  450  lbs. 
due  to  bridge.  Determine  the  maximum  stresses  (both  tensile  and  com- 
pressive) in  the  members  met  by  vertical  planes  immediatelyon  the  right  of 
the  second,  third,  and  fourth  apices  in  the  compression  chord.  Also, 
find  how  many  |-in.  rivets  are  required  to  connect  the  diagonals  met  by 
these  planes  vvitli  the  chords  and  to  prevent  any  tendency  to  longitudinal 
slip  between  the  support  and  tlie  first  apex,  and  between  the  lirst  and 
second  apices  in  the  tension  chord. 

15.  The  accompanying  figure  represents  the  half-truss  for  a  bridge  of 


80  ft.  span.     Show  how  to  determine  the  stresses  ^ 


in  the  several  members.     Depth  at  centre 
ati?=  12  ft.;  at  .,4  =  6  ft. 


12  ft.; 


Fig.  442. 

16.  A  Warren  girder  composed  of  eight  equilateral  triangles  has  its 

upper  chord  in  tension  and  has  every  joint  loaded  with  a  weight  (jf  2  tons, 

the  loads  being  transmitted  to  the  joints  in  the  lower  chord  by  means  of 

vertical  struts.    The  span  =  80  ft.     Find  the  stresses  in  all  the  members. 

Ans.  Bays  in  tension  chord  :         ist  =  5  4/3  ;  2d  =  13  1/3 ; 

3d  =  i8i  1/3  ;  4th  =  21  V3tons. 
Bays  in  compression  chord  :  ist  =  9i  i/3  ;    2d  =  16  ^3 ; 

3d  =  20  v'3  ;  4th  =  21 J^  V3  tons. 
Stresses  in  verticals  :    In  each  vertical  =  2  tons. 


Stresses  in  diagonals : 


1st  =  10  4/3  :  2d  =  8#  4/3 
3d=  7ii^3;   ^th  =  6  4/3; 
5th  =  4f  i/3 ;  ^'th  =  3fr  4^3  : 
7th  =24/3;    8th  =:*  4/3  tons. 

17.  A  Warren  girder,  with  the  platform  on  the  lower  boom,  carries  a 
load  of  20  tons  at  the  centre.  Find  the  stress  in  each  member,  and  also 
find  the  weight  at  each  joint  of  lower  boom  which  will  give  the  same 
stresses  in  the  centre  bays. 

There  are  six  bays  in  the  lower  chord. 

Ans.  Stress  in  each  diagonal  =  ^i  V3  tons. 

Tens,  chord  :  stress  in  ist  bay  =  y  4^3 ;  2d  =  10  4/3  ;    3d  =  V  V'3 

tons. 

Com  p.  chord :  stress  in  ist  bay  =  V  i^3  !  2d  =  V  V3  •    3d  =  20  v'3 

tons. 
Weight  at  each  joint  =5ff  tons. 


it! 


I:  I:  1 


Ml! 


1:  :i.i 

■  1  ■■'   i 

1 

694 


THEORY  OF  STRUCTURES. 


"'-5;Fq^T^q^T^i^]^p=T^p^ 


i8.  The  accompanying  truss  of  240  ft.  span  and  30  ft.  deep  '  5  to  be  de- 
signed for  a  panel  engine  load  of  24,000  lbs.,  a  panel  train  load  of  18,000 
I  lbs.,  and   a   panel   bridge   load   of  12,000 

I  lbs.     Determine  graphically  the  maximum 

stresses  in  the  members  met  by  the  ver- 
tical  MN.     Also,  draw  a  stress  diagram 
Jn  for  the  whole   truss  when   it   is   covered 

Fig.  443.  with  a  uniformly- distributed  live  load  (jf 

180,000  lbs. 

19.  Loads  of  3f ,  6, 6,  6,  and  6  tons  follow  each  other  in  order  over  a  ten- 
panel  truss  at  distances  of  8,  55,  4i,  and  4J  ft.  apart.    Determine  the  posi- 
tion of  the  loads  which  will  give  the  maximum  diagonal  and  chr 
stresses  in  the  third  and  fourth  panels.     Spiin  =  120  ft. 

20.  Determine  the  moment  of  resistance  of  a  floor-beam  for  the  Sault 
Ste.  Marie  Bridge  from  the  following  data:  Floor-beams,  16'  6"  long 
and  23'  io|"  apart ;  the  dead  weight  of  the  flooring,  stringers,  etc. 
=  800  lbs.  per  lineal  foot  of  floor-beam  ;  the  live  load  as  given  in  Fig. 
40/,  Art.  20,  page  639 ;  the  load  is  transmitted  to  the  floor-beam  by  four 
lines  of  stringers  so  spaced  as  to  throw  two-thirds  of  the  load  upon  the 
inner  pair,  which  are  3  ft.  C.  to  C. 

21.  In  a  truss-bridge  the  panels  are  17  ft.  and  the  floor-beams  13  ft. 
in  length.  Loads  of  8,  12,  12,  12,  12,  10,  10,  10,  and  10  tons  follow  each 
other  in  order  over  the  bridge  at  the  distances  of  ^\,  \\,  4},  4^,  7^,  5^,  6^, 
and  5i  ft.  apart.  Determine  the  moment  of  resistance  of  the  beam, 
taking  the  load  due  to  the  platform,  etc.,  to  be  500  lbs.  per  lineal  foot. 

22.  If  the  bridge  in  the  preceding  question  is  of  the  riveted  type  with 
a  single  diagonal  system,  and  with  verticals  at  the  panel  points,  the  num- 
ber of  the  panels  being  ten,  find  how  many  i-in.  rivets  are  required  m 
the  third  panel  from  one  end  to  connect  the  web  with  the  chords,  assum- 
ing the  panel  live  load  to  be  30,000  lbs.  and  the  panel  dead  load  to  be 
10,000  lbs. 

23.  With  the  loading  given  by  Fig.  409,  Art.  20,  page  639,  design  a 
floor-beam  for  a  single-track  bridge  with  panels  22  ft.  long,  the  weight  of 
the  platform  being  450  lbs.  per  square  yard,  and  of  each  longitudinal  200 
lbs.  per  lineal  yard. 

24.  Prepare  a  table  giving  the  stresses  of  the  several  members  of  a 
double-intersection  through-truss  of  154  ft.  span,  20  ft.  depth,  and  with 
eleven  panels.  The  panel  engine,  live,  and  bridge  loads  are  56,000, 
35,000,  and  16,800  lbs.,  respectively. 

25.  Prepare  a  table  giving  the  stresses  in  the  several  members  of  a 
double-intersection  deck-truss  of  342  ft.  span,  33  ft.  depth,  and  with 
eighteen  panels.  (Double-track  bridge.)  The  panel  engine,  train 
(or  live),  and  dead  loads  are  96,000,  54,000,  and  36.200  lbs.,  respectively. 

26.  Prepare  a  table  giving  the  stresses  in  the  several  members  uf  a 


-ill 


EXAMPLES. 


69s 


through-truss  for  a  double-track  bridge  of  342  ft.  span,  40  ft.  depth,  and 
with  nineteen  panels.  The  panel  engine,  live,  and  dead  loads  are  96,000, 
53,000,  and  43,200  lbs.,  respectively  (double-intersection). 


Cl  C3   C3  C4   C5  CO   C?  C3  CO 


Ans. 


ti  ti  ti  ti  td  U  tr  ta  ta  fio 

Fig.  444. 


Chord 
Hanel. 

Mult. 

7326 

Mult. 

5063 

ti  =  t.j 

18 

131868 

'53 

774639 

'. 

—  I 

-  7326 

80J 

40757'* 

'4 

—  I 

7H 

'6 

—  I 

6ii 

'. 

—  I 

53i 

'7 

—  I 

42i 

'8 

—  I 

-  7326 

34i 

174673I 

'. 

—  I 

23* 

'10 

^  I 

( 

'5* 
'53 

fl 

18 

1 

8oi 
7^^ 

Cj 

—  I 

e«l 

fs 

—  I 

53* 

f4 

—  1 

4^t 

f» 

—  1 

34i 

«"« 

—  I 

23* 

f? 

—  1 

'5* 

'■e 

—  I 

4i 

<^» 

—  I 

-  3* 

Total  Shear 

transmitted 

to  Panel. 


906507 
400245* 


•67347* 


Tan- 
gent. 


•45 
•45 
•9 
•9 

•  9 
•9 
•9 
■9 
•9 
■45 
•45 
•9 
•9 
•9 
•9 

•  9 
•9 
•9 
•9 
■9 


Panel  Stress 
from  Shear. 


407929 
180110.475 


150612}^ 


Total 
Panel 
Stress. 


407929 
588039 


921696 


1 

Shear 

1 

Diag. 

Mult. 

5053 

Mult. 

2790 

due  to 
Live 
Load. 

Mult. 

•J274 

Total 
Shear. 

Secant.' 

/> 

18 

'53 

'7' 

1.0965 

'^1 

'7 

63* 

80^ 

1.0965 

rf. 

16 

56* 

72* 

'■345 

d. 

'5 

75795 

48* 

'353'5 

211110 

6'* 

'3985' 

350961 

'■345 

di 

14 

70742 

4'-* 

118575 

'893'7 

53* 

121659 

310976 

'■345 

d. 

«3 

65689 

35* 

99045 

'64734 

42* 

96645 

261379 

'■345 

d. 

12 

60636 

30* 

85095 

•4573' 

34* 

78453 

224184 

'■345 

'^ 

11 

24* 

23* 

'■345 

dn 

10 

2U|^ 

'5* 

••345 

d> 

9 

'.5* 

4* 

'■345 

«I0 

8 

40424 

■2* 

34875 

75299 

-  3* 

-   7959 

'•345 

dn 

7 

3537' 

8* 

237'5 

-M* 

-  32973 

'•345 

Max. 
Stress, 


301528 


90572 
3512a 


7'i  =  139200  lbs. 
«'a  =  35096'  " 
^t  =  3'0976  " 


r'4  =  261379  lbs, 

7'5  =  224184  '• 

''«  =  '77377  " 


7'i  —  142972  lbs. 
vg  =    98955  " 

7',    =      67340 


27.  Prepare  a  table  showing  the  stresses  in  the  several  members 
(including  counters)  of  a  ten-panel  double-track  through  railway  bridge 
of  184^  ft.  span  and  34  ft.  depth,  the  live  and  dead  loads  being  respect- 
ively 2250  lbs.  and  iioo  lbs.  per  lineal  foot.  (Tliamcsville  Bridge.) 

28.  Determine  the  minimum  stress-length  (stress  in  a  member  multi- 
plied by  its  length)  for  a  double-intersection  Pratt  truss  of  1 54  ft.  span  and 
with  eleven  panels.     The  panel  loads  for  engine  =  44,000  lbs.,  for  train  = 


H 


It  1 


696 


THEORY  OF  STRUCTURES. 


27,500  lbs.,  for  bridge  =  13,200  lbs.  ;  coefficientof  working  strengtli  =  8000 
lbs.  per  square  inch  for  both  compression  and  tension. 

29.  A  six-panel  single-intersection  Pratt  truss  is  uniformly  loaded. 
Assuming  the  same  coefficient  of  strength  both  for  compression  and 
tension,  show  that  the  economy  of  material  will  be  greatest  when  tlic 
diagonals  are  inclined  at  32°  25'  to  the  vertical. 

30.  A  double-intersection  truss  for  a  single-track  through-bridge  01 
204  ft.  span  is  29  ft.  deep,  20  ft.  wide,  and  has  twelve  panels.  Find  ihc 
stresses  produced  in  the  members  of  the  leeward  truss  by  a  panel  wind- 
pressure  of  5000  lbs.  acting  8  ft.  above  base  of  rails  (5-ft.  gauge). 

Ans,  Sloping  members  :       1st  =  27500  sec  «  ;       2d  =  1 2708^  sec  a ; 

3d  =  10208^  sec  li\  4th  =  7708^  sec  /i; 
5th  —  52o8i  sec  a  ;  6th  =  2708^  sec  li\ 
7th  =  208^  sec  li. 

Tension  chord  :  ist  panel  =  27500  tan  a  =  2d  ;  3d  =  40208  J  tan  a; 
4th  =  40208^  tan  «  4-  io2o8i  tan  fi; 
5th  =  40208 jr  tan  tr  4.  17916!  tan  /i; 
6ih  =  40208^  tan  nr  +  23125  tan  li. 

Compression  chord  :    ist  =:;  40208^  tan  a  4-  10208^  tan  (i ; 

2d  =  40208^  tan  a  4-  17916^-  tan  (i  ; 

3d  =  40208^-  tan  (f  +  23125  tan  ft  : 
4th  =  402o8i  tan  a  +  25833^^  tan  ft ; 
5th  =  40208^  tan  (t-  -f-  26041 1  tan  ft. 

Verticals:  ist=5ono;      2d  =  7708^;   3d  =  5208!; 

4th  =  2708^ ;  5th  =  2084^  lbs. 
tan  0:  =  ^;     tan  ft  =  f*. 

31.  In  the  preceding  question  find  tiie  maximum  stresses  in  the 
members  of  the  fourth  panel  met  by  a  vertical  plane  ;  engine  panel  Inad 
=  85,000  lbs.,  train  panel  load  =  40,800  lbs.,  bridge  panel  load  =  22,500 
lbs. 

Ans.  Stresses  in  tension  chord  =  456,430.45  lbs. ;  in  compression 
chord  =  645,31 1.77  lbs.  ;  in  sloping  members  =  206,242.5  lbs,; 
and  139,705.62  lbs. 

32.  Each  of  the  two  Pratt  single-intersection  five-panel  trusses  for  a 
single-track  bridge  is  55  ft.  centre  to  centre  of  end  pins  and  1 1  ft.  6  in. 
deep.  Timber  floor-beams  are  laid  upon  the  upper  chords  2^  ft.  centre 
to  centre  ;  the  width  between  the  chords  =  10  ft.  Find  the  proper  scant- 
ling of  the  floor-beams  for  the  loading  given  in  Fig.  407,  page  639. 
Also  determine  the  maximum  chord  and  diagonal  stresses  in  the  centre 
panel  due  to  the  same  live  load. 

33.  Prepare  a  table  giving  the  stresses  in  the  several  members  of  a 
double-intersection  deck-truss  of   342  ft.  span,  40  ft.  depth,  and  with 


EXAMPLES. 


697 


i^ 


B 


nineteen  panels.     (Double-track  bridge.)     The  panel   engine,  train  (or 
live),  iind  dead  loads  are  96,000,  53,000,  and  43,200  lbs.,  respectively. 

34.  Prepare  a  table  giving  the  stresses  in  the  several  members  of  a 
deck-truss  for  a  double-track  bridge  of  342  ft.  span,  33  ft.  depth,  and 
with  eighteen  panels.  The  panel  engine,  live,  and  dead  loads  are  96,000, 
54,000,  and  36,000  lbs.,  respectively. 

35.  The  two  trusses  for  a  16  ft.  roadway  are  each  100  ft.  in  the  clear, 
17  ft.  3  in.  deep,  and  of  the   type   repre-  ^     ^ 
sented  in  the  figure  ;  under  a  live  load  of 
1120  lbs.  per  lineal  foot  the  greatest  total    , 
stress  in  AB  is  35,400  lbs.     Determine  the 
permanent  load.  Fig.  445. 

The  diagonals  and  verticals  are  riveted  to  angle-irons  forming 
part  of  the  flanges.  How  many  f-in.  rivets  are  required  for  the  con- 
nection of  AB  and  BC  at  B?  Also,  how  many  are  required  between 
A  and  C  to  resist  the  tendency  of  the  angle-irons  to  slip  longitudinally  .' 
\Vorking-shear  stress  =  10,000  lbs.  per  square  inch. 

A/is.   708.9  lbs. ;  8  ;  4  ;    II. 

36.  The  compression  chord  of  a  bowstring  truss  is  a  circular  arc  of 
80  ft.  span  and  10  ft.  rise ;  the  bracing  is  of  the  isosceles  type,  the  bases 
of  the  isosceles  triangles  dividing  the  tension  chord  into  eight  equal 
lengths.  Determine  the  maximum  stresses  in  the  members  met  by  a 
vertical  plane  28  ft.  from  one  end.  The  live  and  dead  loads  are  each 
i  ton  per  lineal  foot. 

37.  Design  a  parabolic  bowstring  truss  of  80  ft.  span  and  10  ft.  rise 
for  a  dead  load  of  i  ton  and  a  live  load  of  i  ton  per  lineal  foot.  The 
joints  between  the  web  and  the  tension  chord  are  to  divide  the  latter 
into  eight  equal  divisions. 

38.  The  compression  chord  of  a  bowstring  truss  is  a  circular  arc. 
The  depth  of  the  truss  is  14  ft.  at  the  centre  and  5  ft.  at  each  end  ;  the 
span  =  100  ft.  ;  the  load  upon  the  truss  =  840  lbs.  per  lineal  foot.  Find 
tile  stresses  in  all  the  members.  Determine  also  the  maximum  stresses 
in  the  members  met  by  a  vertical  25  ft.  from  one  end  wlien  a  live  load 
of  1000  lbs.  per  lineal  foot  crosses  the  girder.  What  counter-braces  are 
required  ? 

39.  A  Pratt  truss  with  sloping  end  posts  has  a  length  of  150  ft.  centre 
to  centre,  and  a  height  of  30  ft.  centre  to  centre,  with  panels  15  ft.  long; 
the  dead  load  is  3000  lbs.  per  lineal  foot,  and  the  live  load  1200  lbs. 
Determine  the  maximum  stresses  in  the  end  posts,  in  the  third  post  trom 
one  end,  in  the  middle  of  the  bottom  chord,  and  in  the  members  of  the 
third  panel  met  by  a  vertical  plane. 

40.  Design  a  cross-tie  for  a  double-track  open-web  bridge,  the  ties 


i 


lit      ' 


698 


THEORY  OF  STRUCTURES. 


being  18  ft.  J  in.  centre  to  centre,  nnd  the  live  load  for  the  floor  system 
being  8000  lbs.  per  lineal  foot. 

41.  A  bowstring  roof-truss  of  50  ft.  span,  15  ft.  rise,  and  five  panels  is 
to  be  designed  to  resist  a  wind  blowing  horizontally  with  a  pressure  of 
40  lbs.  per  square  foot.  The  depth  of  the  truss  at  the  centre  is  10  ft. 
Determine,  graphically,  ihe  stresses  in  the  several  members  of  the  truss, 
assuming  that  the  roof  rests  on  rollers  at  the  windward  support. 

42.  A  bowstring  truss  of  120  ft.  spai.  and  15  ft.  rise  V:  of  the  isosceles 
braced  type,  the  bases  of  the  isosceles  triangles  dividing  the  tension 
chord  into  twelve  equal  divisions  ;  the  dead  and  live  loads  are  \  ton  and 
I  ton  per  lineal  foot,  respectively.  Find  the  maximum  stresses  in  ilic 
members  met  by  vertical  planes  immediately  on  the  right  of  the  second 
and  fourth  joints  in  the  tension  chord. 

43.  The  figure  is  a  si<eleton  diagram  of  the  Sault  Ste.  Marie  Bridge 
(C,  P.  R.).     Span  =  239  ft. ;  there  are  ten  panels,  each  of  23.9  ft.,  say  24 


875,000    270,000    410,000    478,000    4»4,00O 

Fig.  446. 
ft. ;  the  length  of  the  end  verticals  =  27  ft.,  of  the  centre  verticals  =  40 
ft.;  width  on  truss  centres  =  i/^-  ft.     The  bridge  is  designed  to  bear  tlie 
loading  given  by  Fig.  407,  page  639.     Show  that — 

{a)  The  stresses  in  every  panel  length  of  each  chord  are  greatest  wiien 
the  third  driver  is  at  a  panel  point ;  and  find  the  value  of  the  several 
stresses. 

{b)  The  stresses  in  the  verticals  a  and  the  diagonals  b  are  greatest 
when  the  third  driver  is  at  a  panel  point ;  and  find  their  values. 

(c)  The  stresses  in  tiie  remaining  members  of  the  truss  are  greatest 
when  the  second  driver  is  at  a  panel  point  ;  and  find  their  values. 

{d)  The  maximum  stresses  in  the  verticals  dv&xy  from  a  tension  of 
64,000  lbs.  to  a  compression  of  1 1,000  lbs. 
(e)  The  stress  in  the  counter-brace  c  is  nil. 

Ans.  The  values  ofthestresses  in  the  several  membersare  marked 

on  the  diagram.     They  are  deduced  from  the  distributions 

given  in  the  table  on  page  642,  and  are  correct  within  a  very 

small  percentage. 

44.  The  figure  represents  a  counterbalanced  s\vinf;-bridi.,^e,  16  ft.  deep 

E  and  wholly  supported  upon  the  turn-table  at  A 


ATwm 


rllWllli 

Fig.  447. 


s     and  B\  the  dead  weight  is  650  lbs.  per  lineal  foot 
of  bridge ;  the  counterpoise  is  hung  from  C  and  A 
Find  its  weight,  assuming  {a)  tliat  the  whole  of  it 
is  transmitted  to  B ;  (b)  that  a  portion  of  it  is  transmitted  to  A  through 


45 

Th 

load 

upor 

foot. 

the 

CD. 

Fine 

so  designe 

upon 

the 

through  th 

points  of  s 

46. 

Fine 

ceding 

que 

closed 

and 

Height  of  ti 

47. 

Prep 

yr^ 

<?2   e 

f/R 

^sN 

/  rA 

1'^  Nj'a' 

k  if 

77T 

27.500,  I 

7.600 

■Pi «  m 


ly 


EXAAfPLES. 


breatest 
rreatest 
hsicin  of 


narkcil 

Ibutions 

a  very 

Jft.  deep 

lie  at./ 

;al  foot 

and  P. 

.\e  of  it 
lb rough 


699 
Also, 


a  member  BE,  sufficient  to  make  the  reactions  at  A  and  5  equal 
determine  the  stresses  in  the  several  members  of  the  truss. 

Ans.  Counterpoise  in  case  (a)  =  26, i62i  lbs. ; 
in  case  (d)  =  22,i86H  lbs. 
Stress  transmitted  through  BE  in  case  {&)=  24,012  lbs. 

45.  The  figure  represents  a  counterbalanced  swing-bridge;  the  dead 
load  upon  the  bridge  is  650  lbs.  per  lineal 
foot ;  the  counterpoise  is  suspended  from 
CD.  Find  its  value,  the  joint  at  E  being 
so  designed  that  the  whole  of  the  load 
upon   the   bridge  is  always  transmitted  fk;.  4,0 

through  the  main  posts  EA,  EB,  and  is  evenly  distributed  between  the 
points  of  support  at  A  and  B. 

Ans.  20,694.3  lbs. 

46.  Find  the  stresses  in  the  several  members  of  the  truss  in  the  pre- 
ceding question  {a)  when  the  bridge  is  open  ;  {b)  when  the  bridge  is 
closed  and  is  subjected  to  a  live  load  of  3000  lbs.  per  lineal  foot. 
Height  of  truss  at  £"  =  16  ft.,  at  i^  =  8  ft. 

47.  Prepare  a  table  giving  the  stresses  in  the  several  members  of  a 
ci   ci  ca  Ci   ch single-intersection  through-truss  of 

20  ft.  depth,  and   with 
lanels.     The   panel  engine. 
Fig.  449.  live,  and  dead  (or  bridge)  loads  are 

27,500,  17,600,  and  8470  lbs.,  respectively. 

Ans. 


til    13   ta    '4     f.'i    'c  .  "^ 


Diag. 

Muit. 

2500 

Mult. 

1600 

Sum.JM 

1 

ult. 

770 

Sum. 

sec. 

Total  Max. 
Stress. 

P 

10 

25000 

45 

72000 

97000' 

55 

42350 

139350;  1.22 

170007 

d. 

9 

■^?^aa 

36 

57600 

80100 

44 

33880 

113980  1.22 

139056 

rf, 

8 

200u'^ 

28 

44800 

64800, 

33 

25410 

90210:  1.22 

110057 

rf» 

7 

17500 

21 

33f^"J 

51100 

22 

16940 

68040  1.22 

83009 

^4 

6 

15000 

'5 

24000 

39000 

11 

8470 

47470,  1.22 

57914 

rf. 

5 

12500 

Ic 

16000 

28500' 

0 

28500I  1.22 

34770 

d^ 

4 

loono 

6 

g6oo 
— 1-  .._ 

19600  — 

II 

-8470 

11130 

1.32 

•3579 

Panel. 

Mult. 

3270 

Mult. 

2370 

Sum. 

tan. 

Panel 
Stress. 

97545 
72366 

54117 
35868 
17619 

Total  Panel 
Stress. 

/,  =  <■, 
'3 

to 

10 

—  1 

—  I 

—  1 

—  1 

32700 

-  3270 

-  3270 

-  3270 

-  3»70 

45 
45 
34 
23 
12 

106650 
106650 
80580 
545'o 
28440 

'39350 

103380 

773'o 

51240 

25170 

7 
10 

97545 
160911 
2240:18 
259896 
2775»5 

1 


I 


700 


THEORY  OF  STRUCTURES. 


Panel. 

Mult. 

3270 

Mult. 

»37«> 

Sum. 

tan. 

Total 
Stress. 

Total  Ma.x.' 
Stress. 

•^i 

lo 

32700 
-  3270 

45  I 
45 

106650  1 
106650 

242730 

f? 

169911 

169911 

c-i 

—  I 

-  3270 

34 

80580 

773'° 

54"7 

224028 

c\ 

—  I 

-  3270 

n 

545'0 

51240 

35868 

259896 

c* 

—  I 

-  3270 

12 

28440 

25170 

17619 

277515 

Ci 

—  I 

-  3270 

1 

2370 

-  900 

—  630 

276885 

i'\  =  35i97o  lbs. ;  i/j  =  90,210 ;  v^  =  68,040 ;  1/4  =  47,470  ;  v^  =  28,500  lbs. 

48.  Compare  the  relative  amounts  of  iron  required  in  the  webs  of  a 
single-  and  a  double-intersection  Pratt  deck-truss  of  loo  ft.  span  and 
having  eight  panels.     Panel  live  load  =  L,  panel  dead  load  =  D. 

49.  The  figure  represents  a  pier,  square  in  plan,  supporting  the  ends 
of  two  deck-trusses,  each  200  ft.  long  and  30  ft.  deep.     The  height  of  tiie 

pier  is  50  ft.  and  is  made  up  of  three  panels,  the 
upper  and  lower  being  each  17  ft.  deep.  Ten 
square  feet  of  bridge  surface  and  ten  square  fett 
of  train  surface  per  lineal  foot  are  subjected  to 
a  wind-pressure  of  40  lbs.  per  square  foot.  Tlie 
centre  of  pressure  for  the  bridge  is  68  ft.,  and 
for  the  train  86  ft.,  above  the  pier's  base.  The 
wind  also  produces  a  horizontal  pressure  of  4000 
Fio.  450.  lbs.  at  each  of  the  intermediate  panel  points  on 

the  windward  side  of  the  pier.     Width  of  pier  =  17  ft.  at  top  and  33$  ft. 

at  bottom.     The  bridge  load  =  1600  lbs.  per  lineal  foot,  live  load  =  3000 

lbs.  per  lineal  foot.     Determine— 

(rt)  The  overturning  moment  (3180  ft.-tons). 

(b)  The  horizontal  force  due  to  the  wind  at  the  top  of  the  pier. 
(61.6  tons.) 

(c)  The  tension  in  the  vertical  anchorage  ties  at  5  and  T.    (Xil.) 
{d)  The  vertical  and  horizontal  reactions  at  T.     (275  and  65.6  tons.) 
Draw  a  diagram  giving  the  wind-stresses  in  all  the  menibers,  and  in- 
dicate which  are  in  tension  and  which  in  compression. 

Ascertain  whether  the  wind-pressure  of  40  lbs.  per  square  foot  upon 
a  train  of  empty  cars  weighing  900  lbs.  per  lineal  foot  will  produce  a 
tension  anywhere  in  the  inclined  posts.  What  will  be  the  tension  in  the 
anchorage  ties  ?     (20.75  tons.) 

Find  the  stresses  in  the  traction  bracing  (i)  when  a  loaded  train  trav- 
elling at  30  miles  an  hour  is  braked  just  as  the  engine  is  over  the  pier 
and  brought  to  rest  in  a  length  of  300  ft.  ;  (2)  when  a  loaded  traih  with 
t'le  engine  over  the  pier  is  started  by  a  sudden  admission  into  the  cylin- 
ders of  steam  at  100  lbs.  per  square  inch.  Stroke  of  cylinder  =  16  in., 
diameter  of  drivers  =  5  ft. 


HI! 


EXAMPLES. 


701 


50.  The  figure  represents  one  half  of  one  of  the  piers  of  the  Bouble 
Viaduct.     The  spans  are  crossed  by  two  lattice-gir-  "^^ 

ders,  14'  9"  deep  and  having  a  deck  platform.  The 
heightof  the  pier  is  183'  9"  and  is  made  up  of  eleven 
panels  of  equal  depth.  Width  of  pier  at  top  =  13' 
li",  at  bottom  =  67' 7".  With  wind-pressure  at 
55.3  lbs.  per  square  foot,  the  total  pressure  on  the 
jjirder,  train,  and  pier  have  been  calculated  to  be  20, 
16.2,  and  20  tons,  acting  at  points  196.2,  210.3,  «i"d 
92.85  ft.,  respectively,  above  the  base.  The  dead 
weight  upon  each  half  pier  is  222J  tons,  of  which  60 
tons  is  weight  of  half  span,  120  tons  the  weight  of 
the  half  pier,  and  42^  tons  the  weight  of  the  train. 
Assuming  that  the  wind-pressure  on  the  pier  is  a 
horizontal  force  of  2  tons  at  each  panel  point  on  the 
windward  side,  and  that  the  weight  of  the  pier  may 
be  considered  as  a  weight  of  6  tons  at  each  panel 
point,  determine — 

(a)  The  overturning  moment.  Fig.  451. 

(b)  The  total  horizontal  force  at  the  top  of  the  pier  due  to  the  wind. 
ic)  The  tension  in  each  of  the  vertical  anchorage  ties  at  5  and  7"  due 

to  the  wind-pressure. 

(d)  The  vertical  and  horizontal  reactions  at  T. 
Show  tliat  the  greatest  compressive  stress  occurs  in  the  member  RT, 
and  that  it  amounts  to  422  tons. 

Draw  a  stress  diagram  giving  the  stresses  in  all  the  members,  indi- 
caiin;^  which  are  in  tension  and  which  in  compression.  Width  of  pier 
at  ^  =  20  ft.,  ax.  B  =  23^  ft.,  at  C  =  36^  ft. 

What  will  be  the  effect  of  braking  the  train  when  running  at  30  miles 
an  hour,  so  as  to  bring  it  to  rest  within  a  distance  of  220  ft.  ?  Width 
of  pier  in  direction  of  bridge  =  9I  ft.  at  top  and  =  20  ft.  at  bottom. 

Ans.—{a)  9188  ft.-tons;  (b)  39.9  tons;  {c)  24^  tons,      {d)  Hori- 
zontal reaction  =  59.9  tons  ;  vertical  reaction  =  247  tons. 
51.  The  accompanying  figure  represents  a  portion  of  a  cantilever  truss, 
the  horizontal  distances  of  the  points  A,  B,  C 
from  the  free  end  being  h  ,  h ,  h,  respectively. 
The  boom  ABC  is  inclined  at  an  angle  a,  and  the 
boom  A'  VZ  at  an  angle  ft,  to  the  horizon.     Find 
the  deflections  at  the  end  of  the  cantilever  due  to 
X  Y  2.    {a)  an  increase  kAB  in  the  length  of  AB;  (2)  an 

Fig.  45a.  increase  k^B  K  in  the  length  of  ^  F ;  (3)  a  decrease 

kiXY  in  the  length  oi  XY;  (4)  a  decrease  iiBX  in  the  length  of  BX. 

k.hAB 
^'"-^'^  BX^^nABX' 


M 


Si   ij 


«li 


li!'L-':». 


702 


THEORY  OF  STRUCTURES. 


(3) 


^•^sin  i^^K 


.    (  BX cos  a        ,  ,         „,,,, 
<4)  >f'<  i    .       ,  „  V.  -  />(cot  BXY~  cot  yiZ.'.Vi 
(  sill  ABA  ' 

In  the  preceding  question,  if  ki  =  ^-a  =  /'a  =  ^'<  =  l\  and  if  ^/  W  is 
parallel  to  ^5^,  and  AX  to  i?  F.  show  that  the  angle  between  WX  and 
^K  after  deformation 

=  2/C-(cot  ABX  +  cot  /y  F.V). 

Hence  also,  if  the  truss  is  of  uniform  deptli  d,  show  that  the  "  deviation  " 

2k 

of  the  boom  per  unit  of  length  is  constant  and  equal  to  ^• 

a 

52.  Six  bars  have  to  be  arranged   upon  a  steel  pin  ;  each  bar  is  i  in. 

wide  and  is  subjected  to  a  stress  of  64,000  lbs.     Should  the  bars  be  ar- 


M.OOO  lb8..4— ^ 


U.OOO-^ 


MjOOO-*- 


M,0004 
81.000  ■<—^^~ 
W,000< ^ 


"t:^ — ^>-61,U001b». 


2- 


->- 64,000 


'J  >um 


Fig.  433.— Method  1. 


;2s— »-M,ooo 
7 >-64,000 


J 


■^ — ►.M.OOO 


Fig.  454. — Method  2. 

ranged  according  to  method  i  or  method  2  .'    Why  ?    Determine  the  di- 
ameter of  the  pin. 

53.  The  accompanying  sketch  represents  one  of  the  pin  connections 
in  a  certain  bridge  which  was  recently  overthrown.  The  two  innermost 
bars  are  web  members  inclin-jd  to  the  horizon  at  an  angle  whose  cosine 


'V^ra^ 


48,000  lba<'pl>4'hnri  3'!        | 

I  '  t>^  — y^8.100  lb8, 

«,«00  \\mt^V4-  i        I 


I  I  ' 


Fig.  455. 


is  .815.  The  thickness  of  the  bars  and  the  maximum  stresses  to  which 
they  are  severally  subjected  are  shown  on  the  diagram.  Is  the  3-in. 
wrought-iron  pin  sufficiently  strong? 


CHAPTER   XII. 


SUSPENSION-BRIDGES. 


1.  Cables. — The  modern  suspension-bridge  consists  of  two 
or  more  cables  from  which  the  platform  is  suspended  by  iron 
or  steel  rods.  The  cables  pass  over  lofty  supports  (piers),  and 
are  secured  to  anchorages  upon  which  they  exert  a  direct  pull. 

Chain  or  link  cables  are  the  most  common  in  England  and 
Europe,  and  consist  of  iron  or  steel  links  set  on  edge  and 
pinned  together.  Formerly  the  links  were  made  by  welding 
the  heads  to  a  flat  bar,  but  they  are  now  invariably  rolled  in 
one  piece,  and  the  proportional  dimensions  of  the  head,  which 
in  the  old  bridges  are  very  imperfect,  have  been  much  im- 
proved. 

Hoop-iron  cables  have  been  u.sed  in  a  few  cases,  but  the 
practice  is  now  abandoned,  on  account  of  the  difficulty  attend- 
ing the  manufacture  of  endless  hoop-iron. 

Wire-rope  cables  are  the  most  common  in  America,  and 
form  the  strongest  ties  in  proportion  to  their  weight.  They 
consist  of  a  number  of  parallel  wire  ropes  or  strands,  compactly 
bound  together  in  a  cylindrical  bundle  by  a  wire  wound  round 
the  outside.  There  are  usually  seven  strands,  one  forming  a 
core  round  which  are  placed  the  remaining  six.  It  was  found 
impossible  to  employ  a  seven-strand  cable  in  the  construction 
of  the  East  River  Bridge,  New  York,  as  the  individual  strands 
would  have  been  far  too  bulky  to  manipulate.  The  same  ob- 
jection held  against  a  thirteen-strand  cable  (thirteen  is  the  next 
number  giving  an  approximately  cylindrical  shape),  and  it  was 
finally  decided  to  make  the  cable  with  nineteen  strands.  Seven 
of  these  are  pressed  together  so  as  to  form  a  centre  core,  around 
which  are  placed  the  remaining  twelve,  the  whole  being  con- 
tinuously wrapped  with  wire. 

703 


\m 


704 


THEORY  OF  STRUCTURES. 


In  laying  up  a  cable  yreat  care  is  required  to  distribute  the 
tension  uniformly  amon<,'st  the  wires.  This  ma\-  be  effected 
either  by  giving  each  wire  the  same  deflection  or  by  usiit^ 
straight  wire,  i.e.,  wire  which  when  unrolled  upon  the  floor 
from  a  coil  remains  straight  and  shows  no  tendency  to  spriii;' 
back.  The  distribution  of  stress  is  practically  uniform  in  un- 
twisted wire  ropes.  Such  ropes  are  spun  from  the  wires  and 
strands  without  giving  any  twist  to  individual  wires. 

The  back-stay  is  the  portion  of  the  cable  extending  from  an 
anchorage  to  the  nearest  pier. 

The  elevation  of  the  cables  should  be  sufficient  to  allow  fur 
settling,  which  chiefly  arises  from  the  deflection  due  to  the  load 
and  from  changes  of  temperature. 

The  cables  may  be  protected  from  atmospheric  influence 
by  giving  them  a  thorough  coating  of  paint,  oil,  or  varnish,  but 
wherever  they  are  subject  to  saline  influence,  zinc  seems  to  be 
the  only  certain  safeguard. 

2.  Anchorage,  Anchorage  Chains,  Saddles. — The  an- 
chorage, or  abutment,  is  a  heavy  mass  of  masonry  or  natural 
rock  to  which  the  end  of  a  cable  is  made  fast,  and  which  re- 
sists by  its  dead  weight  the  pull  upon  the  cable. 


Fig.  456.  Fig.  457.  Fig.  458. 

The  cable  traverses  the  anchorage  as  in  Figs.  456  to  458, 
and  passes  through  a  strong,  heavy  cast-iron  anchor-plate,  and. 
if  made  of  wire  rope,  has  its  end  effectively  secured  by  turning 
it  round  a  dead-eye  and  splicing  it  to  itself.  Much  care,  how- 
ever, is  required  to  prevent  a  wire-rope  cable  from  rustin"  on 
account  of  the  great  extent  of  its  surface,  and  it  is  co;  • 
advisable  that  the  wire  portion  of  the  cable  should  alv  .r- 

minate  at  the  entrance  to  the  anchorage  and  there  be  a.  lied 
to  a  massive  chain  of  bars,  which  is  continued  to  the  anclior- 
plate  or  plates  and  secured  by  bolts,  wedges,  or  keys. 


ANCHORAGE,  ANCHORAGE   CHAINS,  SADDLES. 


705 


In  order  to  reduce  as  much  as  possible  the  depth  to  which 
it  is  necessary  to  sink  the  anchor-plates,  the  anchor-chains  are 
frequently  curved  as  in  Fig.  458.  This  gives  rise  to  an  oblitiue 
force,  and  the  masonry  in  the  part  of  the  abutment  subjected 
to  such  force  should  be  laid  with  its  beds  perpendicular  to  the 
line  of  thrust. 

The  anchor-chains  arc  made  of  compound  links  consisting 
alternately  of  an  odd  and  an  even  number  of  bars.  The  friction 
of  the  link-heads  on  the  knuckle-plates  considerably  lessens 
the  stress  in  a  chain,  and  it  is  therefore  usual  to  diminish  its 
sectional  area  gradually  from  the  entrance  E  to  the  anchor. 
This  is  effected  in  the  Niagara  Suspension  Bridge  by  varying 
the  section  of  the  bars,  and  in  the  East  River  Bridge  by  vary- 
ing both  the  section  and  the  number  of  the  bars. 

The  necessity  of  preserving  the  anchor-chains  from  rust  is 
of  such  importance  that  many  engineers  consider  it  most 
essential  that  the  passages  and  channels  containing  the  chains 
and  fastenings  should  be  accessible  for  periodical  examination, 
painting,  and  repairs.  This  is  unnecessary  if  the  chains  are 
first  chemically  cleaned  and  then  embedded  in  good  hydraulic 
cement,  as  they  will  thus  be  perfectly  protected  from  all  at- 
mospheric influence. 

The  direction  of  an  anchor-chain  is  changed  by  means  of  a 
saddle  or  knuckle-plate,  which  should  be  capable  of  sliding  to 
an  extent  sufificient  to  allow  for  the  expansion  and  contraction 
of  the  chain.  This  may  be  accomplished  without  the  aid  of 
rollers  by  bedding  the  saddle  upon  a  four-  or  five-inch  thickness 
of  asphalted  felt. 

The  chain,  where  it  passes  over  the  piers,  rests  on  saddles, 
the  object  of  which  is  to  furnish 
bearings  with  easy  vertical  curves. 
''  it^her  the  saddle  may  be  constructed 
a>>  in  Fig.  459,  so  as  to  allow  the 
cable  to  slip  over  it  with  compara- 
ti'  y  little  friction,  or  the  chain  may  be  secured  to  the  saddle, 
and  the  saddle  supported  upon  rollers  which  work  over  a  per- 
fectly true  and  horizontal  bed  formed  by  a  saddle-plate  fixed 
to  the  pier. 


1 


■a 


m 


1. 


'■::  r 


7o6 


THEORY  OF  STRUCTURES. 


3.  Suspenders. — The   suspenders   are  the  vertical  or  in. 
clined  rods  which  carry  the  platform. 


Fig.  460. 


Pig.  461. 


Fig.  463. 


Fig.  463. 


In  Fig.  460  the  suspender  rests  in  the  groove  of  a  cast- 
iron  yoke  which  straddles  the  cable.  Fig.  461  shows  the 
suspender  bolted  to  a  wrought-iron  or  steel  ring  which  em- 
braces the  cable.  When  there  are  more  than  two  cables  in 
the  same  vertical  plane,  various  methods  are  adopted  to  insure 
the  uniform  distribution  of  the  load  amongst  the  set.  In  Fig. 
462,  for  example,  the  suspender  is  fastened  to  the  centre  of  a 
small  wrought-iron  lever  PQ,  and  the  ends  of  the  lever  are 
connected  with  the  cables  by  the  equally  strained  rods  PR 
and  QS.  In  the  Chelsea  bridge  the  distribution  is  made  by 
means  of  an  irregularly  shaped  plate  (Fig.  463),  one  angle  of 
which  is  supported  by  a  joint-pin,  while  a  pin  also  passes 
through  another  angle  and  rests  upon  one  of  the  chains. 

The  suspenders  carry  the  ends  of  the  cross-girders  (floor- 
beams),  and  are  spaced  from  5  to  20  ft.  apart.  They  should 
be  provided  with  wrought-iron  screw-boxes  for  purposes  of 
adjustment. 

4.  Curve  of  Cable. — .  ASEA.  An  arbitrarily  loaded  flexible 
cable  takes  the  shape  of  one  of  the  catenaries,  but  the  true 
catenary  is  the  curve  in  which  a  cable  of  uniform  section  and 
material  hangs  under  its  own  v/eight  only. 

Let  A  be  the  lowest  point  of  the  cable,  and  take  the  ver- 
tical through  A  as  the  axis  oi  y. 

Take  the  horizontal  tlirough  0  as  the  axis  of  x,  the  origin 
O  being  chosen  so  that 


p  .AO  =  H  =  mp. 


(0 


/  being  the  weight  of  a  unit  of  length  of  the  cable,  and  //' the 
horizontal  pull  at  A. 


tflil! 


ciblc 
true 
ami 

Ycr- 

[rigin 

the 


CURVE   OF  CABLE. 


707 


m  OX  AO  \s  the  parameter,  or  modulus,  of  the  catenary,  and 
OG  is  the  directrix. 

Let  X,  y  be  the  co-ordinates  of  any  point  P,  the  length  of 
the  arc  AP  being  s. 

Draw  the  tangent  /'T'and  the  ordinate  PN. 

The  triangle  PNT  is  evidently  a  triangle  of  forces  for  the 
portion  AP^  PN  representing  the  weight  of  AP  (viz.,  ps),  PT 


the  tangential  pull  T  at  P,  and  NT  the  horizontal  pull  H  at 
A. 

,,±=:,,nPTN=g^=.^=^,      ...    (2) 
ax  TN       H      m  ^ 

which  gives  the  differential  equation  to  the  catenary. 
It  may  be  easily  integrated  as  follows : 


^s  -      I     ,   fdyV         /     ,    s'         I 


(3) 


or 


(is 


Vs'  4-  m' 


dx 
m' 


,\\og{s-\-Vs'-\-m')  = 


w; 


£  being  a  constant  of  integration. 

When  x  =  0,    s  —  o,     and  therefore     log  m  =  c. 
Hence, 


log 


J  +  V/  4-  vt^     X 


m 


m' 


7o8 


THEORY  OF  STRUCTURES. 


.# 


or 


s-\-^/s^  -\-m*=me"', 


m 


Again, 
and  hence, 


j=__(^«_^-«.). (^y 


dy       s       I ,   *         -  5. 
-j-  =  —  —  -U  >"  —  e    '«); 
ax      m      2  ' 


7  =  -(^  "  +  ^  ") (5) 


The  constant  of  integration  is  zero,  since  y  =  m  when 
X  =0. 

The  last  equation  is  the  equation  to  the  catenary,  while  eq. 
(4)  gives  the  length  of  the  arc  AP. 

By  equations  (4)  and  (5), 


f  =  s'  +  m\ 


(6) 


Draw  NM  perpendicular  to  PT^  and  let  the  angle  PTN  = 
PNM  -  e.    Then 


PM  =  PN  sinewy 


and 


MN  =  PNcos  e=y        '"  - 

i^s'  +  in 


\=5,   ,     .     .    (7) 
=  w,  .     .    .    (8) 


since  tan  6/  =  -f-  =  _. 
ax      in 

Thus,  the  triangle  PMN  possesses  the  property  that  the 
side  PM  is  equal  to  the  length  of  the  arc  AP,  and  the  side 
J/iV  is  equal  to  the  modulus  ;«(=  ^(9). 

The  area  APNO 


ydx  =  — (f  ••  —  e~  m):=z  ms  =  2  X  triangl 


gPMN. 


The 


PG  bein^ 

At  A, 

Again 


P.  being  tl 
These 
and  constr 
the  assuni] 
in  practice 
density. 

Case  I 
posed  of  a 


tween  the  Ic 

any  given  lii 
the  horizonti 


CURVE  OF  CABLE, 
The  radius  of  curvature,  p,  at  P 


709 


1-  +  (I)T    (£)■     y    ^ 

d  y  y  m 


dx' 


m 


PG  being  perpendicular  to  PT. 

At  A,  y  =.  m,  and  the  radius  of  curvature  is  also  m. 


(9) 


(10) 


Again, 


ps 


PT_ 

NP 


y 

cosec  6  =  -, 


.'.T  =  py', 


(II) 


H  =  pm=pp,',       (12) 

p^  being  the  radius  of  curvature  at  A. 

These  catenary  formul.'e  are  of  little  if  any  use  in  the  design 
and  construction  of  suspension-bridges,  as  they  are  based  upon 
the  assumption  of  a  purely  theoretical  load  which  never  occurs 
in  practice,  viz.,  the  weight  of  a  chain  of  uniform  section  and 
density. 

Case  B.  Let  the  platform  be  suspended  from  chains  com- 
posed of  a  number  of  links,  and  let  W  be  the  whole  weight  be- 


tween the  lowest  point  O  of  the  chain  and  the  upper  end  P  oi 
any  given  link.  Let  the  direction  of  this  link  intersect  that  of 
the  horizontal  pull  {H)  at  O  in  E.    Drop  the  perpendicular  PN^, 


7IO 


THEORY  OF  STRUCTURES. 


The  triangle  PNE  is  evidently  a  triangle  of  forces ;  and  if  the 

angle  PEN  =  6, 

^      PN       W 
tan  e 


and  hence 


NE 
tan  e  (X  W. 


H' 


Thus,  by  treating  each  link  separately,  commencing  with  the 
lowest,  the  exact  curve  of  the  chain  may  be  easily  traced. 

Generally  speaking,  the  distribution  of  the  load  may  be 
assumed  to  be  approximately  uniform  per  horizontal  unit  of 
length,  the  load  being  suspended  from  a  number  of  points 
along  each  chain  or  cable  by  means  of  rods.  The  curve  of  the 
cable  will  then  be  a  parabola. 

Let  w  be  the  intensity  of  the  load  per  horizontal  unit  of 
length. 

Let  X,  y  be  the  co-ordinates  of  any  point  P  of  the  cable 
with  respect  to  the  horizontal  OX  and  the  vertical  6^  F  as  axes 
of  X  and  y,  respectively. 

Let  ^  be  the  inclination  of  the  tangent  at  /'to  the  horizon- 
tal. The  portion  OP  of  the  cable  is  kept  in  equilibrium  by 
the  horizontal  pull  H  at  0,  by  the  tangential  pull  T  at  P,  and 
by  the  load  ivx  upon  OP,  which  acts  vertically  through  the 
middle  point  E  of  ON,  PN  being  the  ordinate  at  P. 

Hence,  the  tangent  at  P  must  also  pass  through  E,  and 
PEN  is  a  triangle  of  forces.     Hence, 


X 

wx  ~~  y' 


,        2H 

or    x^  =  — -y, 


(I) 


the  equation  to  a  parabola  with  its  vertex  at  0,  its  axis  vertical, 

,        2H 

and  its  parameter  equal  to  — -. 


Again, 


I 


T_PE^_ 

H~  EN~  cos  e* 


and  hen 


and  the , 

that  at  t 

Also, 


The 


so  that  th 


5-  Par 

and  B,  res 
Let  01 
By  equ 


/ 


Denote  th< 


Also, 


PARAMETER,  ETC. 


and  hence 


Tcose  =  H  = 


wx 


2y' 


7" 


(2) 


and  the  horizontal  pull  at  every  point  of  the  cable  is  the  same  as 
that  at  the  lowest  point. 
Also, 


^        wx  „  WX' 

T  — sec  u  =  — 

2y  2y 


/iH — ^-='wx  K     \-\ -, 

\J         -  V         4/ 


The  radius  of  curvature  at  P 


2y[ 
x'\~ 

Hi 

so 

that  the  radius 

at  Ois 
or 

Po  = 

H 

w' 

« 

H  = 

wp^. 

w 
H 


5.  Parameter,  etc. — Let  h^,  //,  be  the  elevations  of  A 
and  B,  respectively,  above  the  horizontal  line  COD,  Fig.  465. 
Let  OD  =  a, ,  OC  —  a^,  and  let  a,  +  a,  =  a  =  CI). 
By  equation  (i),  Art.  4. 


/2H  _    a,    _ 

y    w   ~  ^h\  ~ 


a. 


a,-\-a. 


a 


Vh,        Vh,       Vh,  +  Vh,       Vh,-\-Vh, 
Denote  the  parameter  by  P.     Then 


w 


Also, 


tan  d 


\Vh,  +  VhJ' 

_'2y  _  ZUX  _  24r  _  Fy 

~~x~ll~~P~^\l  P 


i 


'ill 


712  THEORY  OF  STRUCTURES. 

If  ^, ,  6*,  be  the  values  of  ^  at  /i  and  B,  respectively. 


'-sll 


tan  6^, 


Note.—\{  h.^K  —  hy 


and     tan  d. 


v/^ 


and  hence 


a  a" 


tan  e,  =  —  =  tan  ^, . 


6.  Length  of  Arc  of  Cable.— Let  OP  =  s,  Fig.  465. 


Since    tan  6  = 


7l>  W 

sec'  ^^^  =  -jjdx  =.  -jyds  cos  6^, 


or 


ds 


H   dd 


w  cos'  6' 


Hence, 

_Hf'de         H 

111    VQ 


w  ^°   cos'  B      2W 


—  I  tan  ^  sec  6*  +  log,  (tan  ^  +  sec  ^)  j . 


Again, 


tan  (y  =  -fyx, 


and 


/■ 


w 


sec^  =  ^/i+— 4r». 


WEIGHT  OF  CABLE. 


7ti 


Note. — An  approximate  value  of  the  length  of  the  arc  may 
be  obtainf^H  as  follows : 

ds^  =  dx^^df  =  dx-^  I  I  +  (g)'[  =  dx^  (i  +  -^). 

/  I   'W^X'^\ 

.',  ds  =  dx\i  -\ /TT J.  approximately. 

Integrating  between  0  and  P, 
s^OP  =  x+~ 


I    IV^X* 


6  //"   ~     ^  2>x' 


7.  Weight  of  Cable. — The  ultimate  tenacity  of  iron  wire 
is  90,000  lbs.  per  square  inch,  while  that  of  steel  rises  to 
200,000  lbs.,  and  even  more.  The  strength  and  gauge  of  cable 
wire  may  be  insured  by  specifying  that  the  wire  is  to  have  a 
certain  ultimate  tenacity  and  elastic  limit,  and  that  a  given 
number  of  lineal  feet  of  wire  is  to  weigh  one  pound.  Each  of 
the  wires  for  the  cables  of  the  East  River  Bridge  was  to  have 
an  ultimate  tenacity  of  3400  lbs.,  an  elastic  limit  of  1600  lbs., 
and  14  lineal  feet  of  the  wire  were  to  weigh  one  pound.  A  very 
uniform  wire,  having  a  coefificient  of  elasticity  of  29,000,000  lbs., 
has  been  the  result,  and  the  process  of  straightejiing  has  raised 
the  ultimate  tenacity  and  elastic  limit  nearly  8  per  cent. 

Let  W^  be  the  weight  of  a  length  «,  (=  OD)  of  a  cable  of 
sufficient  sectional  area  to  bear  safely  the  horizontal  tension  H. 

Let  fF,  be  the  weight  of  the  length  J,  (  =  OA)  of  the  cable 
of  a  sectional  area  sufficient  to  bear  safely  the  tension  7",  at  .^. 

Let /be  the  safe  inch-stress. 

Let  q  be  the  specific  weight  of  the  cable  material. 
Then 


and 


W,=  y  a,q 


W^  =  ^L^s^. 


f 


llfil 


'0 


I 


/H 


THEORY  OF  STRUCTURES. 


•••'^.=  "'.^".,  =  ^(.  +  f^)(.+f4-...), 


or 


^F,=  fr.(i+^-^),  nearly. 


A  saving  may  be  effected  by  proportioning  any  given  section 

to  the  pull  across  that  section. 

At  any  point  {x,  y)  the  pull  =  H  sec  B,  and  the  correspond. 

//  sec  /9 
ing  sectional  area  =  7 — .     The  weight  per  unit  of  length 


Hs&cd 


/ 


q,  and  the  total  weight  of  the  length  j,  (=  OA)  is 


But  x'  =  -j-v. 


<  J 


Hence, 


and  also 


■■■"■-'in 

Hq(      .   aK\  , 


The  weight  of  a  cubic  inch  of  steel  averages  .283  lb. 

The  weight  of  a  cubic  inch  of  wrought-iron  averages  .2781b. 

IT 

The  volume  in  inches  of  the  cable  of  weight  W^,=  \2afj . 


DEFLECTION  OF  CABLE. 


W, 


1 2a, 


jj=.  .283  lb.  or  .278  lb., 
7 


7»5 


":if 

according  as  the  cable  is  made  of  steel  or  iron. 

Let  the  safe  inch-stress  of  steel  wire  be  taken  at  33,960  lbs., 
of  the  best  cable-iron  at  14,958  lbs.,  and  of  the  best  chain-links 
at  9972  lbs.     Then 

W,  =  Ha,  X  .283  X  -^  =  -—-  for  steel  cables  ; 

339CX)       lOOOO 

W,  =  Ha,  X  .278  X -^  =  -~   for  iron  cables ; 

'  14958      4500 

W,  =  Ha,  X  .278  X =  — --    for  link  cables. 

'         '  9972         3000 

NoU. — About  one-eighth  may  be  added  to  the  net  weight  of 
a  chain-cable  for  eyes  and  fastenings. 

8.  Deflection  of  a  Cable  due  to  an  Elementary  Change 
in  its  Length. 

By  the  corollary  of  Art.  6  the  total  length  {S)  of  the  cable 
AOB  is 

Now  a,  and  a^  are  constant ;  /i,  —  //,  is  also  constant,  and 
therefore  d/i,  =  dh^.     Hence, 


If  the  alteration  in  length  is  due  to  a  change  of  t°  in  the 
temperature, 

dS  =  ctS, 

c  being  the  coefificient  of  linear  expansion  and  =  5 — -— — -^ 
per  degree  Fahr.  for  wrought-iron. 


m  i 


Li 


7i6 


THEORY  OF  STRUCTURES. 


In  England  the  effective  range  of  temperature  is  about  60° 
Fahr.,  while  in  other  countries  it  is  usual  to  provide  for  a  range 
of  from  100°  to  150°  F, 

If  the  alteration  is  due  to  a  pull  of  intensity /per  unit  of 
area, 

dS  =  jj  S, 
E  being  the  coefficient  of  elasticity  of  the  cable  material. 

If //,=./!,  =  //, 

a  \G  h 

a,  =  a,  =  —  ,    and     dS  = d/t. 

'       2'  3  rt 

9.  Curve  of  Cable  from  which  the  load  is  suspended  by  a 
series  of  sloping  rods. 

/y 


T'       0         E 

Fig.  466. 

Let  0  be  the  lowest  point  of  such  a  cable.  Let  the  tangent 
at  O,  and  a  line  through  O  parallel  to  the  suspenders,  be  the 
axes  of  X  and  }>,  respectively. 

Let  tc'  be  the  intensity  of  the  oblique  load.  Consider  a 
portion  OP  of  the  cable,  and  let  the  co-ordinates  of /'with 
respect  to  OX,  OVhe  x  and  j. 

Draw  the  ordinate  FN,  and  let  the  tangent  at  Pmeet  ON 
in  E. 

As  before,  PNE  is  a  triangle  of  forces,  and  E  is  the  middle 
point  of  ON.     Then 

w'x      PN      2y  .       2H 


H  ~  NE 


or 


w 


-y^ 


the  equation  to  a  parabola  with  its  axis  parallel  to  (9F  and  its 

2// 
focus  at  a  point  S,  where  i\SO  —  — >  . 


CURVE   OF  CABLE   WITH   OBLIQUE  SUSPENDERS.        /I/ 

Cor.  I.  Let  the  axis  meet  the  tangent  at  O  in  T\  and  let 
its  inclination  to  OX  be  /'. 

Let  A  be  the  vertex,  and  ON'  a  perpendicular  to  the  axis. 
Then 

SO  =  ST'  =  SA-i-AT'  =  SA  +  AN'. 

But    4AS  .  AN'  =  ON"  =  N' T"  tzin' i=  4AN "  tan' i. 

AS 
.:AS=  AN'  tan' t,    and     SO  =  AS(i  4-  cot' t)  =  -.^r-- 

^     '  '      sin  z 


Hence, 


the  parameter  =  dtAS  =  4^6^  sin'  /. 


Cor.  2.  Let  P  be  the  oblique  load  upon  the  cable  between 
0  and  P. 

Let  Q  be  the  total  thrust  upon  the  platform  at  E. 

"     w  "     "    load  per  horizontal  unit  of  length. 

"     q    "     "    rate  of  increase  of  thrust  along  platform. 

"     t     "     •'    length  of  PE. 

Then 

w 

. ,     and     q  ■=■  %v  cot  i ; 


w 


sm  I 


*      3 

w  X 


H  =  — "-  =  2w'.  SO  =  2AS4^-.  =  2AS-^.; 
2y  sm  t  sm'  i 

X         y 

x'' 
f  =  y"^  -\-  ^ — \-  xy  cos  i. 
4 

Cor.  3.  Let  s  be  the  length  of  OP,  and  let  6*  be  the  inclina- 
tion of  PE  to  (9  F.     Then 

s  =  AP-  AO 

=  ^^^^'  I  tan  (90°  -  ^)  sec  (90°  -  0) 

+  log,  I  tan  (90°  -  ^)  +  sec  (90°  -ft)\-  tan  (90°-  i)  sec  (90° -z) 

—  log,{  tan  (90°-  /)  4-  sec  (90°  -  /) }  | 

fffiin'ti         .  „  ,.  .,,      cot  &  4-  cosec  0  \ 

= 7—  <  cot  ^  cosec  6  —  cot  t  cosec  ?  4-  log, — -^—. — .  f . 

2w       i  "  cot  ^ -|*  cosec  ?  ) 


isi; 


'•■!?«- B' 


7i8 


THEORY  OF  STRUCTURES. 


It  may  be  easily  shown,  as  in  the  Note  to  Art.  6,  that  ap- 
proximately 

.  ,    2    y  sin'  i 
3  ^+7  cos* 

10.  Pressure  upon  Piers,  etc. 

Let  T^  be  the  tension  in  the  main  cable  at  A. 

"     T;  "     "  "        "     "    back-stay  at  A. 

"    a,  ft  be  the  inclinations  to  the  horizontal  of  the  tangents 
at  A  to  the  main  cable  and  back-stay,  respectively. 
The  total  vertical  pressure  upon  the  pier  at  A 

=  T',  sin  a  -f-  7",  sin  /?  =  R. 

The  total  resultant  horizontal  force  at  A 

=x  Tj  cos  «  ~  T",  cos  ft  —  Q. 

If  the  cable  is  secured  to  a  saddle  which  is  free  to  move 
horizontally  on  the  top  of  the  pier  (Fig.  467), 

Q  —  the  frictional  resistance  to  the  tendency  to  motion, 
or       Qt  f^A 
Ml  being  the  corresponding  coefficient  of  friction. 


Fig.  467. 

Let  D,  Fig.  468,  be  the  total  height  of  the  pier,  and  let  IV 
be  its  weight. 

Let  FG  be  the  base  of  the  pier,  and  K  the  limiting  position 
of  the  centre  of  pressure. 


^ 


AUXILIARY  OR   STIFFENING    TRUSS. 


719 


Let  /,  q  be  the  distance  of  P  and  W,  respectively,  from  K. 
Then 

for  stabiltty  of  position  Q  _ tt , 

and  for  stability  of  friction,  when  the  pier  is  of  masonry, 

^  the  coefficient  of  friction  of  the  masonry. 


P+  W 


If  jw,  is  sufficiently  small  to  be  disregarded,  Q  is  approxi- 
mately nil,  and  Z',  cos  i*  =  T,  cos  fi  =■  H.  The  pressure  upon 
the  pier  is  now  wholly  vertical  and  is  =  ^(tan  a  -\-  tan  (i). 

When  the  cable  slides  over  smooth  rounded  saddles  (Fig. 
459),  the  tensions  T^  and  T^  are  approximately  the  same. 

Thus, 

R  =  T^(s\x\  Of  +  sin  /?)     and     Q  =  T^cos  a  —  cos  /3). 

U  a  =  fi,  Q  =  O,  and  the  pressure  upon  the  pier  is  wholly 
vertical,  its  amount  being  2  7",  sin  a. 

The  piers  are  made  of  timber,  iron,  steel,  or  masonry,  and 
allow  of  great  scope  in  architectural  design. 

The  cable  should  in  no  case  be  rigidly  attached  to  the  pier, 
unless  the  lower  end  of  the  latter  is  free  to  revolve  throutrh  a 
small  angle  about  a  horizontal  axis. 

II.  Auxiliary  or  Stiffening  Truss. — The  object  of  a  stiff- 
ening truss  (Fig.  469)  is  to  distribute  a  passing  load  over  the 
cable  in  such  a  manner  that  it  cannot  be  ilistorted.  The  pull 
upon  each  suspender  must  therefore  be  the  same,  and  this  vir- 


tually assumes  that  the  effect  of  the  extensibility  of  the  cable 
and  suspenders  upon  the  figure  of  the  stiffening  truss  may  be 


disregarded. 


-e*^* 


^     Si'-r  '■ 


720 


THEORY  OF  STRUCTURES. 


The  ends'6>  and  A  must  be  anchored,  or  held  down  by 
pir.s,  but  should  be  free  to  move  horizontally. 

Let  there  be  n  suspenders  dividing  the  span  into  {fi  ^  i) 
equal  segments  of  length  a. 

Let  P  be  the  total  weight  transmitted  to  the  cable,  and  z 
the  distance  of  its  centre  of  gravity  from  the  vertical  through  0. 

Let  T  be  the  pull  upon  each  suspender. 

Taking  moments  about  O, 

n(n  -4-  i^  nl 

Ps=  r(«  +  2a  +  3«  +  .  .  .  +  nd)  -  Ta   '    ^   '  =  T—, 

2  2 

/being  the  length  of  OA. 

Also,  if  t  is  the  intensiy  of  pull  per  unit  of  span, 


tl  =  nT,     and  hence     P2  =  t —  . 

2 


Let  there  be  a  central  suspender  of  length  s.     There  will, 
11  —  I 


therefore,  be 


suspenders  on  each  side  of  the  centre. 


The  parameter  of  the  parabola  =  ~t  • 
Hence,  the  total  length  of  all  the  suspenders 

..+2|«-^.+4^.{/+2«+3-F...i-(^y]l 


'       /        24  \     '    3  « -|~  J 


If  there  is  no  central  suspender,  i.e.,  if  n  is  even, 

/        h      n    \ 
the  total  length  =z  {n  —  \)\s  -\ TT]' 

Denote  the  total  length  of  suspenders  by  L.    Then 

the  strcss-leni^th  =  TL  ==  -,PL. 

^  nl 


'il    ;. 


i       ■ 


AUXILIARY  OR   STIFFENING    TRUSS. 


721 


Let  tu  be  the  uniform  intensity  of  the  dead  load. 

Case  I.    T/u  bridge  partially  loaded. 

Let  w'  be  the  maximum  uniform  intensity  of  the  live  load, 
and  let  this  load  advance  from  A  and  cover  a  length  AB. 

Let  OB  =  X,  and  let  /?, ,  R^  be  the  pressures  at  O  and  A, 
respectively. 

For  equilibrium, 

R,  +  R,  +  tl-wl-w'{l-x)  =  0)  .     .     .     (I) 

/'  /'       za' 

R^lJ^t--w~--{l-x)^      =0.    .     .     .     (2) 


*  g 


Also,  since  the  whole  of  the  weight  is  to  be  transmitted 
through  the  suspenders, 


i/z=wl-{-  w\l  —  x). 
From  eqs,  (i),  (2),  and  (3), 


(3) 


w  X 


-R.  =  —j{l-x)  =  R,, (4) 


which  shows  that  the  reactions  at  0  and  A  are  equal  in  mag- 
nitude but  opposite  in  kind.     They  are  evidently  greatest  when 

/ 
X  =  -,  i.e.,  when  the  live  load  covers  half  the  bridge,  and  the 

w'l  * 

common  value  is  then  -x-  . 

The  shearing  force  at  any  point  between  0  and  B  distant  x' 
from  0 

=  R^J^{t-7V).x'  -^w'^-^[x'  --^,.      .      .      (5) 


i?i>  Ei 


''i'i- 

which  becomes 


W  X 

T7 


(/  -  X) 


R^  =  /?,  when   x'  equal  x. 


Thus  the  shear  at  the  head  of  the  live  load  is  equal  in  magni- 
tude to  the  reaction  at  each  end,  and  is  an  absolute  maximum 


III  i! 


722 


THEORY  OF  STRUCTURES. 


when  the  live  load  covers  half  the  bridge.     The  web  of  the 

w'l 
truss  must  therefore  be  designed  to  bear  a  shear  of  -^-  at  the 

o 

centre  and  ends. 

Again,  the  bending  mome  it  at  any  point  between  0  and  B 

distant  x'  from  O 


„    ,  ,    t  —  w    ,„       w'  I  —  X 
R.x'A x"  =  — 

'         '  2  2 


—{x'^-xx'),     .    .    (6) 


which  is  greatest  when  ;jr'  z=— ,  i.e.,  at  the  centre  of  OB,  its 

w'  I  —  X 
value  then  being  —  --  — j-x''.     Thus,  the  bending  moment  is 

d  .  2 

an  absolute  maximum  when  -r-ilx''  —  x^)  =  o,  i.e.,  when  x  =  -/, 

dx  y         '        '  2  T 

and  its  value  is  then P. 

54 

The  bending  moment  at  any  point  between  B  and  A  dis- 
tant x'  from  O 


t  —  IV 


w 


W    X, 


=  R,x'  +  -^-x'^  -  -{x'  -xy  =  -  j{x'  -  x){l-  x'\  (7) 


which  is  greatest  when  -r--,\{x'  —  x){l  —  .r')}  =  o,  i.e.,  when 

l^x 
x'  =: ,   or   at    the    centre    of  AB,  its   value    then    being 

*w  X 
q"  -(/—  xy.     Thus,  the  bending  moment  is  an  absolute  maxi- 

mum  when  -,-|;r(/  —  xy\  =  o,  i.e.,  when  x  —  -,  and  its  value  is 
ax  3 

za' 
then  +  — /'. 
54 

Hence,  t/ie  maximum  baiding  moments  of  the  unloaded  and  \ 
loaded  divisions  of  the  truss  are  equal  in  magnitude  but  opposite 
in  direction,  and  occur  a.'  the  points  of  trisection  {D,  C)  of  OA ; 


f%:k«ii 


•m'' 


I' 


i  i  I 


AUXILIARY  OR   STIFFENING    TRUSS. 


723 


■when  the  live  load  covers  one-tliird  ( 4C)  and  two-thirds  {AD)  of 
the  bridge,  respectively. 

Each  chord  must  evidently  be  designed  to  resist  both 
tension  and  compression,  and  in  order  to  avoid  unnecessary 
nicety  of  calculation,  the  section  of  the  truss  may  be  kept  uni- 
form throughout  the  middle  half  of  its  length. 

Case  II.  A  single  concentrated  load  W  at  any  point  B  of 
the  truss.  W^now  takes  the  place  of  the  live  load  of  intensity 
w' . 

The  remainder  of  the  notation  and  the  method  of  pro- 
cedure being  precisely  the  same  as  before,  the  corresponding 
equations  are 


R.  +  K  4-  (^  -w)l-W  =  o. 


•       •       •       • 


(!') 


lir 


/  —  w 
RJ+  —r-l' 


W{l-x)  =  o (2') 


W 
t-w=-j. 


(30 


-R^  =  ~{x~^=R, (40 


which  shows  that  the  reactions  at  O  and  A  are  equal  in  mag- 
nitude but  opposite  in  kind.  They  are  greatest  when  x  ~  o 
and  when  x  =  I,  i.e.,  when  IV  is  either  at  O  or  at  A,  and  the 

common  value  is  then  — . 

2 

The  shearing  force  at  any  point  between  O  and  B  distant  x' 

from  O 


W I  l\ 

^R^^{t-W)x'  =  -j{x'-X^-;^,    . 


'vhich  is  a  maximum  when  x'  =  x,  and  its  value  is  then 


(50 


W 


w 


The  web  must  therefore  be  designed  to  bear  a  shear  of 
throughout  the  whole  length  of  the  truss. 


1' 

1 

.  ; 

>«'\''' 


,  ;''! 

1 

k 

i 

1 

724 


THEORY  OF  STRUCTURES. 


Again,  the  bending  moment  at  any  point  between  0  and 
B  distant  x'  from  O 


W  (-tr" 


^R^,'^it-w)-^-j^-^x\--x)\.    •     (6' 


First,  let  w  <—.     The  bending  moment  is  positive  and  is  a 
maximum  when  x'  =  x,  its  value  then  being 

W 


Next,\QX.  X  >— .     The  bending  moment  is  then  neg-ativc and 

I 
is  a  maximum  when  x'  =^  x  —  —,  its  value  then  being 


W( 


'-I 


The  bending  moment  at  any  point  between  B  and  A  dis. 
tant  x'  from  O 

r"  W  Ix'        \ 

=  R^^'  +  (^  _  ^)i_  _  W{x'  -x)=  -j{x'  -  l)[  -^  -  .-),  (7') 


which  is  a  maximum  when 


i.e.,  when  x'  =:  x  -{-  -,  and  its  value  is  then jlx 1  . 

Note. — The  stiffening  truss  is  most  effective  in  its  action,  but 
adds  considerably  to  the  weight  and  cost  of  the  whole  struc- 
ture. Provision  has  to  be  made  both  for  the  extra  truss  and 
for  the  e.Ktra  material  required  in  the  cable  to  carry  this  extra 
load. 


'!        1 


AUXILIARY  OR   STIFFENING   TRUSS. 


725 


Stiffening  Truss  hinged  at  the  Centre. — Provision  may  be 
made  for  counteracting  the  straining  due  to  changes  of  tem- 
perature by  hinging  the  truss  at  the  centre  E. 

Let  a  live  load  of  intensity  tv'  advance  from  A. 

First,  let  the  live  load  cover  a  length  AB  =^  x  [>  — 

Let  R^ ,  Rj  be  the  pressures  at  O,  A,  respectively. 
The  equations  of  equilibrium  are 


^.  +  ^,  +  (^  -  w)/  -W'X  =  0',    .     . 


(0 


IV 


^4 +  ('-'")?-¥■"  =  °' 


(3) 


Eqs.  (2)  and  (3)  being  obtained  by  taking  moments  about  E. 
Hence,  ■'■■■■■' 

zv' 
i  -.w  =  —  -jt{1^  -  4^^  +  2x') ;      ...     (4) 


I  w 


R,  =  -^{r--4^x+3x^)',  ....    (5) 


I  w' 

R.-^-jil-x)\    . 


...    (6) 


Next,  let  the  live  load  cover  the  length  BO  (  <  -j. 

Let  AB  =  ;r  as  before,  and  let  R,',  R,',  t'  be  the  new  values 
of  R,,  R,,  t,  respectively. 

The  equations  of  equilibrium  are  now 


R,'  +  R,'  +  (/'  -  wy  -  zoy  ^  x)  =  o; 


.    (7) 


7- 

2 


w 


K'\  -f  it'  -  «')  8  ~  7^^^  -  •^)  =  o  ; 


(8) 


726 


THEORY  OF  STJiUCTURES. 

i 


R/--\■{t'-^vY^^o^ 


and  hence, 


w 


(9) 


t'  -w=^  2y,-(/  -  ;r)'  [=  -  (/  -  w  -  w')-\  ;     .     (10) 


I  w 


^:=--^-'f{i'-Alx-\-lx^){=-Ry,    .    (II) 


r:  = 


I  w 


j{i-xy{=-R,) (12) 


Diagravi  of  Maximum  Shearing  Force. — The  shear  at  any 
point  distant  z  from  A  in  the  unloaded  portion  BO  when  the 
hve  load  covers  AB 

=  R.-V{t-w){l-z) (13) 

=  -  \R,'-\- it'  -w-  w'){l -  z)\ 

=  -   {R,'-{-{t'  -  w){/-2)  -  w'{l-2)\ 

=  minus  the  shear  at  the  same  point  when  AB  is 
unloaded  and  the  live  load  covers  BO. 

For  a  given  value  of  z  the  maximum  shear,  positive  or 
negative,  at  any  point  of  OB,  is  found  by  making  (see  eq.  (13) ) 

dR,-\-{l-~2)d{t-w)  =  o, 


or 


w 


tif , 


_-(_  2/+  IX)  -jAl-  2){-  4/+  4^)  =  O, 
or  «f 


x  =  l 


4^—2/ 

AZ-l' 


(14) 


and  occur 
^B^  and  i 


"-«-  ">'  e,s.  (,),  (;,,  (,3j_  (,^^^ 


727 


Fig.  4;^j. 


the  ;«^4r.W^  ,/,,^^^  ^  i^^'^fljlif 
and  mav  hp  r^r,  ^  ~  x'    '    '     05) 

:Mrf/4'.'^,:~';/,,''rti.e  ordinate      .j; 

the  shears  are  greatest  when 

"=^      ^''  U        u 

and  their  values  ar. 

-^'ues  are,  respectively, 

^^gain,  the  shear  -.f 

=  »«wtl,e  shear  at  th.  , 

»W.^  and  the  ,.Ve  irarret'l^;^"  ^^  '^ 

the  shears  I  i!?',, 
increasing  for  a  given  v,l„„    , 

"-™u„  whe„ : :7'^'-;^  ^  With  /_  .,  ,„,,  ,,^^^,^^_^^  ^ 

the  maximum  shear  =  i  i^'/, 

^--=:^^L-/rCt^--  -overs 

^''nen  It  covers  BO.     It 


(16) 


728 


THEORY  OF  STRUCTURES. 


may  be  represented  by  the  ordinate  {^positive  or  negative)  of 
the  curve  orsq. 

For  example,  at  the  points  denned  by 

z^x^l,  \l,  1/,  \l,  \l, 

the  maximum  shears  given  by  eq.  (i8)  are,  respectively. 

Diagram  of  Maxinuun  Bending  Moment. — The  bending 
moment  at  any  point  in  BO  distant  z  from  A  when  the  live 
load  covers  AB 


=  Ril  -  z)  -\-{t-  w)- 


{i-zr 


(19) 


t'    —  7t)   —    W/  '1^^ ~     (. 


=  -   I  RXl  -  .)  +  {f  -  u^jtlJL  _  Jl^  \ 

■=  minus  the  bending  moment  at  the  same   point  when 
the  live  load  covers  BO. 
Hence,  by  eqs.  (4),  (5),  (19),  the  bending  moment 


I  lU 


I  zv 


=  ±  - -yi/"  -  V'tr  +  3,v')(/  -z)  T  -  -j{l'  -  4/-V  +  2x\l  -z) 
For  a  given  value  of  z  this  is  a  maximum  and  equal  to 

2lz 


w'  zl  ~  zl  —  2:. 

^  T        (/  -  2Z) 


when      X  = 


/-f-  2Z' 


Thus,  the  maximum  bending  moment  may  be  represented 
by  the  ordinate  {positive  or  negative)  of  a  curve. 
For  example,  at  the  points  defined  by 


/, 


lA 


i/, 


|/. 


2' 


AUXILIARY  OR   STIFFENING    TRUSS. 


729 


the  bending  moments  •  x-c  greatest  when  x  = 


*/; 


their  values  being,  respectively, 

o,     TM^w'r,     T-.hf'l',      ^sh^'P.      o- 


-h \ 


^- -■-' 


Fic.  471. 

The  absolute  maximum  bending  moment  may  be  found  as 
follows : 

For  a  given  value  of  x  the  bending  moment  (see  eq.  (19) )  is 
a  maximum  when 


R^J^{t-w){l-  z)^0, 


or 


I-  Z-- 


R. 


t  —  w 
Hence,  the  maximum  bending  moment 


^  2  t  —  IV       ^8     /'  ~  4/^  +  ix" 


It  will  be  an  absolute  maximum  for  a  value  of  x  found  by  put 
ting  its  differential  with  respect  to  x  equal  to  nil. 
This  differential  easily  reduces  to 

ix'  -  gix^  +  erx  -r  =  0. 

X  =  ^l  is  an  approximate  solution  of  this  equation,  and  the  cor- 
responding maximum  bending  moment  :=  ^^f-j^t'V. 

The  preceding  calculations  show  f/iat  at  every  point  in  its 
length  the  truss  may  be  subjected  to  equal  maximum  shears  and 
equal  maximum  bending  moments  of  opposite  signs. 

Again,  it  may  be  easily  shown,  in  a  similar  manner,  tiiat 


ill! 


\  Hi 


V:   ■     \ 


il!i1»f*sii 


730 


THEORY  OF  STRUCTURES. 


when  a  single  weight  W  travels  over  the  truss, 

the  maximum  positive  shear  at  a  distance  z  from  A 

W 

the  maximum  negative  shear 

W 
either  =  -^(Z"  —  %lz-{-  4^') 

I   W 
or  =  -  -j{ll  -  4^) ; 

and  the  maximum  bending  moment 


=  ±  Y2{/  -  Z){1  —  2Z). 


12.  Suspension-bridge  Loads. — The  heaviest  distributed 
load  to  which  a  highway  bridge  may  be  subjected  is  that  due 
to  a  dense  crowd  of  people,  and  is  fixed  by  modern  French 
practice  at  82  lbs.  per  square  foot.  Probably,  however,  it  is 
unsafe  to  estimate  the  load  at  less  than  fromicxD  to  140  lbs.  per 
square  foot,  while  allowance  has  also  to  be  made  for  the  con- 
centration upon  a  single  wheel  of  as  much  as  36,000  lbs.,  and 
perhaps  more. 

A  moderate  force  repeatedly  applied  will,  if  the  interval 
between  the  blows  corresponds  to  the  vibration  interval  of  the 
chain,  rapidly  produce  an  excessive  oscillation  (Chap.  Ill, 
Cor.  2,  Art.  24).  Thus,  a  procession  marching  in  step  across 
a  suspension-bridge  may  strain  it  far  more  intensely  than  a 
dead  load,  and  will  set  up  a  synchronous  vibration  which  may 
prove  absolutely  dangerous.  For  a  like  reason  the  wind 
usually  sets  up  a  wave-motion  from  end  to  end  of  a  bridge. 

The  fixctor  of  safety  for  the  dead  load  of  a  suspension-biidge 
should  not  be  less  than  2^  or  3,  and  for  the  live  load  it  is 
advisable  to  make  it  6.  With  respect  to  this  point  it  maybe 
remarked  that  the  efficiency  of  a  cable  does  not  depend  so 
much  upon  its  ultimate  strength  as  upon  its  limit  of  elasticity, 


MODIFICATIONS  OF   THE  SIMPLE   SUSPENSION-BHIDGE.    731 

and  so  long  as  the  latter  is  not  exceeded  the  cable  remains  un- 
injured. For  example,  the  breaking  xveight  of  one  of  the  1 5-inch 
cables  of  the  East  River  Bridge  is  estimated  to  be  12,000  tons, 
its  limit  of  elasticity  being  81 18  tons ;  so  that  with  ii^  only  as  a 
factor  of  safety,  the  stress  would  still  fall  below  the  elastic 
limit  and  have  no  injurious  effect.  The  continual  application 
of  such  a  load  would  doubtless  ultimately  lead  to  the  destruc- 
tion of  the  bridge. 

The  dip  of  the  cable  of  a  suspension-bridge  usually  varies 
from  iV  *°  tV  °^  ^^^  span,  and  is  rarely  as  much  as  ji-j,  except 
for  small  spans.  Although  a  greater  ratio  of  dip  to  span  would 
give  increased  economy  and  an  increased  limiting  span,  the 
passage  of  a  live  load  would  be  accompanied  by  a  greater  dis- 
tortion of  the  chains  and  a  larger  oscillatory  movement. 
Steadiness  is  therefore  secured  at  the  cost  of  economy  by 
adopting  a  comparatively  flat  curve  for  the  chains. 

13.  Modifications  of  the  Simple  Suspension-bridge. — 
The  disadvantages  connected  with  suspension-bridges  are  very 
great.  The  position  of  the  platform  is  restricted,  massive 
anchorages  and  piers  are  generally  required,  and  any  change  in 
the  distribution  of  the  load  produces  a  sensible  deformation  in 
the  structure.  Owing  to  the  want  of  rigidity,  a  considerable 
vertical  and  horizontal  oscillatory  motion  may  be  caused,  and 
many  efforts  have  been  made  to  modify  the  bridge  in  such  a 
manner  as  to  neutralize  the  tendency  to  oscillation. 

{ci)  The  simplest  improvement  is  that  shown  in  Fig.  472, 
where  the  point  of  the  cable  most  liable  to  deformation  is 
attached  to  the  piers  by  short  straight  chains  AB. 


Fio. 


472. 


(^)  A  series  of  inclined  stays,  or  iron  ropes,  radiating  from 
the  pier-saddles,  may  be  made  to  support  the  platform  at  a 
number  of  equidistant  points  (Fig.  473).  Such  ropes  were  used 
in  the  Niagara  Bridge,  and  still  more  recently  in  the  East  River 


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23  WEST  MAIN  STREET 

WEBSTER,  N.Y.  14580 

(716)  872-4503 


732 


THEORY  OF  STRUCTURES. 


Bridge.  The  lower  ends  of  the  ropes  are  generally  made  fast 
to  the  top  or  bottom  chord  of  the  bridge-truss,  so  that  the  cor- 
responding chord  stress  is  increased  and  the  neutral  axis  pro- 
portionately  displaced.  To  remedy  this,  it  has  been  proposed 
to  connect  the  ropes  with  a  horizontal  tie  coincident  in  position 
with  the  neutral  axis.    Again,  the  cables  of  the  Niagara  and 


Fig. 


473- 


East  River  bridges  do  not  hang  in  vertical  planes,  but  are  in- 
clined inwards,  the  distance  between  them  being  greatest  at 
the  piers  and  least  at  the  centre  of  the  span.  This  drawing  in 
adds  greatly  to  the  lateral  stability,  which  may  be  still  further 
increased  by  a  series  of  horizontal  ties. 

{c)  In  Fig.  474  two  cables  in  the  same  vertical  plane  arc 
diagonally  braced  together.    In  principle  this  method  is  similar 


Fig. 


474- 


to  that  adopted  in  the  stiffening  truss  (discussed  in  Art.  1 1),  but 
is  probably  less  efficient  on  account  of  the  flexible  character  of 
the  cables,  although  a  slight  economy  of  material  might  doubt- 
less be  realized.  The  braces  act  both  as  struts  and  ties,  and 
the  stresses  to  which  they  are  subjected  may  be  easily  calcu- 
lated. 

{d)  In  Fig.  475  a  single  chain  is  diagonally  braced  to  the 
platform.    The  weight  of  the  bridge  must  be  sufficient  to  insure 


Fig.  475. 


that  no  suspender  will  be  subjected  to  a  thrust,  or  the  efficiency 
of  the  arrangement  is  destroyed.     An  objection  to  this  as  well 


II' 


MODIFICATIOXS  OF   THE   SIMPLE   SUSPENSIOX-BRIDGE.   733 

as  to  the  preceding  method  is  that  the  variation  in  the  curva- 
ture of  the  chain  under  changes  of  temperature  tends  to  loosen 
and  strain  the  joints. 

The  principle  has  been  adopted  (Fig.  476)  with  greater  per- 
fection in  the  construction  of  a  foot-bridge  at  Frankfort.    The 


Fig.  476. 


girder  is  cut  at  the  centre,  the  chain  is  hinged,  and  the  rigidity 
is  obtained  by  m*  ans  of  vertical  and  inclined  braces  which  act 
both  as  struts  and  ties. 

{e)  In  Fig.  477  the  girder  is  supported  at  several  points  by 


Fig.  477. 


straight  chains  running  directly  to  the  pier-saddles,  and  the 
chains  are  kept  in  place  by  being  hung  from  a  curved  chain  by 
vertical  rods. 

(/)  It  has  been  proposed  to  employ  a  stiff  inverted  arched 
rib  of  wrought-iron  instead  of  the  flexible  cable.  All  straining 
action  may  be  eliminated  by  hinging  the  rib  at  the  centre  and 
piers,  and  the  theory  of  the  stresses  developed  in  this  tension 
rib  is  precisely  similar  to  that  of  the  arched  rib,  except  that 
the  stresses  are  reversed  in  kind. 

{g)  The  platform  of  every  suspension-bridge  should  be 
braced  horizontally.  The  floor-beams  are  sometimes  laid  on 
the  skew  in  order  that  the  two  ends  of  a  beam  may  be  sus- 
pended from  points  which  do  not  oscillate  concordantly,  and 
also  to  distribute  the  load  over  a  greater  length  of  cable. 


r-r% 


734 


THEORY  OF  STRUCTURES, 


EXAMPLES. 

1.  The  span  of  a  suspension-bridge  is  200  ft.,  the  dip  of  the  chains  is 
80  ft.,  and  the  weight  of  the  roadway  is  i  ton  per  foot  run.  Find  the  ten- 
sions at  the  middle  and  ends  of  each  chain.  Ans.  31^  tons ;  58.94  tons. 

2.  Assuming  that  a  steel  rope  (or  a  single  wire)  will  bear  a  tension  of 
15  tons  per  square  inch,  sliow  that  it  will  safely  bear  its  own  weight  over 
a  span  of  about  one  mile,  the  dip  being  one-fourteenth  of  the  span. 

Ans,  Max.  tension  =  33,074  lbs. 

3.  Show  that  a  steel  rope  of  the  best  quality,  with  a  dip  of  one-seventh 
of  the  span,  will  not  break  until  the  span  exceeds  7  miles,  the  ultimate 
strength  of  the  rope  being  60  tons  per  square  inch. 

Ans.  Max.  tension  =  59.545  tons  per  square  inch. 

4.  The  river  span  of  a  suspension-bridge  is  930  ft.  and  weighs  5976 
tons,  of  which  1439  tons  are  borne  by  stays  radiating  from  the  summit 
of  each  pier,  while  the  remaining  weight  is  distributed  between  four 
15-in.  steel-wire  cables,  producing  in  each  at  the  piers  a  tension  of  2064 
tons,    rind  the  dip  of  the  cables.  Ans.  66.44  ft- 

The  estimated  maximum  traffic  upon  the  river  span  is  131 1  tons 
uniformly  distributed.     Determine  the  increased  stress  in  the  cables. 

Ans.  596.4  tons. 
To  what  extent  might  the  traffic  be  safely  increased,  the  limit  of 
elasticity  of  a  cable  being  81 16  tons,  and  its  breaking  stress  i:,3oo  tons  ? 

Ans.  To  13,303  tons  uniformly  distributee!. 


5.  If  the  span  =  /,  the  total  uniform  load  =  W,  and  the  dip  = 


/ 


\i 


show  that  the  maximum  tension  =  1.58  ff'',  the  minimum  tension 
=  1.5  \V,  the  length  of  the  chain  =  1.018/,  and  find  the  increase  of  dip 
corresponding  to  an  elongation  of  i  in.  in  the  chain. 

6.  A  cable  weighing  /  lbs.  per  lineal  foot  of  length  is  stretched  be- 
tween supports  in  the  same  horizontal  line  and  20  ft.  apart.  If  the  ma.\- 
imum  deflection  is  \  ft.,  determine  tiie  greatest  and  least  tensions. 

Ans.  Parameter  ;//  =  100  ft.;  max.  tension  =  100^/;    min.  ten- 
sion =  100/. 

7.  A  light  suspension-bridge  carries  a  foot-path  8  ft.  wide  over  a 
river  90  ft.  wide  by  means  of  eight  equidistant  suspending  rods,  the  dip 
being  10  ft.  Each  cable  consists  of  nine  straight  links.  Find  their  several 
lengths.     If  the  load  upon  the  platform  is  120  lbs.  per  square  foot,  and 


EXAMPLES. 


735 


if  one-fourth  of  the  load  is  borne  by  the  piers,  find  the  sectional  areas 
of  the  several  links,  allowing  10,000  lbs.  per  square  inch. 

Ans,  Lengths  in  ft.,  10;    10.049;  10.198;  10.44;  'o??- 

Tensions  in  lbs.,  45000;  4500^^101  ;  45001/104;  4 500 if/ 1 09  ; 

4500  1/116. 
Areas  in  sq.  in.,  4.5  ;  4.522  ;  4.59  ;  4.698  ;  4.847. 

8.  A  suspension-bridge  of  200  ft.  span  and  20  ft.  dip  has  48  sus- 
penders on  each  side;  the  dead  weight  =  3000  lbs.  per  lineal  foot ;  the 
live  load  =  2000  lbs.  per  lineal  foot.  Find  the  maximum  pull  on  a  sus- 
pender, the  maximum  bending  moment  and  the  maximum  shear  on  the 
stiffening  truss.  Also,  find  the  elongation  in  the  chain  due  to  the  live  load. 

Atts.  Max.  pull  =  12,500  lbs.;  max.  shear  =  30,000  lbs.;  max. 
B.M.  =  i,o66,666S  ft. -lbs. ;  elongation  =  89,600,000  -*-  EA,  A 
being  sectional  area  of  a  cable,  and  E  the  coefficient  of  elas- 
ticity. 

9.  A  foot-path  8  ft.  wide  is  to  be  carried  over  a  river  100  ft.  wide  by 
two  cables  of  uniform  sectional  area  and  having  a  dip  of  10  ft.  Assum- 
ing the  load  on  the  platform  to  be  112  lbs.  per  square  foot,  find  the 
greatest  pull  on  tiie  cables,  tlieir  sectional  area,  length,  and  weight. 
(Safe  stress  =  8960  lbs.  per  square  inch  ;  specific  weight  of  cable  =  4S0 
lbs.  per  cubic  foot.) 

Ans.  H  =  —:--  7^=56,000  lbs.;  area  =6.73  sq.  in.; 
V29 

length  =  io2|  ft.;  weight  =  ■y.^oz.ti,  lbs. 

10.  Find  the  depression  in  the  cables  in  the  last  question  due  to  an 
increment  of  length  under  a  change  of  60°  F.  from  the  mean  temperature. 
(Coefficient  of  expansion  =  1  -*-  144000.)  Ans.  .0S02  ft. 

11.  Each  side  of  the  platform  of  a  suspension-bridge  for  a  span  of  100 
ft.  is  carried  by  nine  equidistant  suspenders.  Design  a  stitl'eiiing  truss  for 
a  live  load  of  1000  lbs.  per  lineal  foot,  and  determine  the  pull  upon  the 
suspenders  due  to  the  live  load  when  the  load  produces  (i)  an  ixhsoliitc 
ma.ximuin  shear ;  (2)  an  absolute  tiiaximum  beiidiit^  moment. 

Ans.  Max.  shear  =  6250  lbs.;  max.  B.M.  =  92, 592^ J  ft. -lbs.;  pull 
on  suspender  =  (i)  2777J  lbs.,  (2)  =  18515?  "'«•  »•■  yT^ZW  lbs. 

12.  In  a  suspension-bridge  (recently  blown  down)  each  cable  was  de- 
signed to  carry  a  total  load  of  84  tons  (including  its  own  weight).  The 
distance  between  the  piers  =  1270  ft.;  the  deflection  of  the  cable  =  91  ft. 
Find  (a)  the  length  of  the  cable ;  (/')  the  pull  on  the  cable  at  the  piers 
und  at  the  lowest  point ;  (c)  the  amounts  by  which  these  pulls  are  changed 
by  a  variation  of  40°  F.  from  the  mean  temperature;  (d)  the  tension  in 
the  back-stays,  assuming  them  to  be  approximately  straight  and  inclined 
to  the  vertical  at  the  angle  whose  tangent  is  \. 


'%     i 


ill 


\k. 

1 

736 


THEORY  OF  STRUCTURES. 


Ans.—(a)  1287.4ft.;  (6)  //= =  1461^3  tons;  (r) depression  due 

1.04 

tochaiigeof  lenip.  =  .936  ft.  and  amount  of  change  in  //=  '-     // 

y 

=  li  tons,  in    7=1.45  tona ;    ('/)  394-55    tons,  neglecting   pier 
friction, 

13.  The  platform  of  the  bridge  in  tiie  preceding  question  was  liun^^ 
from  the  cables  by  means  of  480  suspenders  (240  on  each  aide).  Find 
the  pull  on  each  suspender  and  the  total  length  of  the  suspenders,  the 
lowest  point  of  a  cable  being  14  ft.  above  tiie  platform. 

Arts.  .35  ton;  10,565^]',  ft. 

14.  A  suspension-bridge  has  a  dip  of  10  ft.  and  a  span  of  300  ft.  Fimi 
the  increase  of  dip  due  to  a  change  of  100"  F.  from  the  mean  tempera- 
ture, the  coefficient  of  expansion  being  .00125  P*^*"  '80°  F, 

Aus.  1. 17  ft. 
Also,  find   the  corresponding   flange   stress  in  the  stifTeniiig  truss. 
which  is  I2i  ft.  deep,  the  coeiricient  of  elasticity  being  8000  tons. 

A/ts.  6.24  tons. 

15.  The  ends  of  a  cable  are  attached  to  saddles  free  to  move  hori/^on- 
tally.  If  Jrt  is  the  horizontal  movement  of  each  saddle  due  to  the  ex- 
pansion of  the  cables  in  the  side  spans,  and  if  ^S  is  the  extension  of  the 
chain  between  the  two  saddles,  show  that  the  increment  cf  the  dip  (//)  is 
approximately 

(3a  _/r 

{8  A       a 

16.  The  platform  of  a  suspension-bridge  of  150  ft.  span  is  suspended 
from  the  two  cables  by  88  vertical  rods  (44  on  each  side)  ;  the  dip  of  the 
cables  is  15  ft.;  there  are  two  stiffening  trusses  ;  the  dead  weight  is  2J40 
lbs.  per  lineal  foot,  of  which  o/tf-Ad// is  divided  equally  between  the  two 
piers.  Find  the  stresses  at  tiie  middle  and  ends  of  the  cables  when  a 
uniformly  distributed  load  of  78,750  lbs.  covers  one  half  of  the  bridge. 
.-\lso,  find  the  maxiinun\  shears  and  bending  moments  to  which  the  stid- 
ening  trusses  are  subjected  when  a  live  load  of  1050  lbs.  per  lineal  foot 
crosses  the  bridge. 

Ans.  Pull  on  suspender  =  2741 J  lbs. ;  // =    -_^  T=  203,4374  lbs. 

y29 

Max.  shear  on  each  truss  at  centre  and  due  to  78,750  lbs. 
=  9843^  lbs.  =  that  due  to  1050  lbs.  per  lineal  foot. 

Max.  H.  M.  due  to  78,750  lbs.  is  at  centre  of  loaded  and  un- 
loaded halves  and  =  184,570/^  ft. -lbs. 

Abs.  max.  B.M.  due  to  1050  lbs.  per  lineal  foot  is  at  points 
of  trisection  and  =  218,750  ft. -lbs. 


3  '' 
16  A 


EXAMPLES. 


717 


17.  Solve  the  preceding  question  when  the  trusses  are  hinged  at  the 
centre. 

Ans.  Pull  on  suspender  =  2741 S  lbs.  ;   // =  -  _  7^  =  1 54,2 18}  lbs. 

Max.  shear  due  to  78,750  lbs.  =  9843J  lbs.  at  centre  of 
span  and  at  end  of  loaded  half  of  bridge ;  max.  shears 
due  to  1050  lbs.'  per  lineal  foot  =13,125,  5906J,  4921  J, 
8305  ,"2^,  and  9843^  lbs.  at  ends  of  the  half  truss  and  at 
the  points  dividing  the  half  span  into  four  equal  seg- 
ments. 

Max.  H.  M.  due  to  78,750  lbs.  is  at  centre  of  half  truss  and 
=  i84,57o,V  ft.-lbs.  Max.  B.  M.  due  to  1050  lbs.  per  lineal 
foot  =176,1803 S3,  221,4841,  and  1 53,808 J |!  ft.-lbs.  at  points 
dividing  the  half  truss  into  four  equal  segments. 

18.  Show  that  tiic  total  extension  of  a  cable  of  uniform  sectional  area 
A  under  a  uniformly  distributed  load  ai  inteiisiiy  7*/  is 


JoP_[ 


I  + 


6  ^'\ 
3    rl' 


16 
3 


/  being  the  span  and  d  the  dip. 

19.  The  dead  weight  of  a  suspension-bridge  of  1600  ft.  span  is  i  ton 


per  lineal  foot;  the  dip  = 


span 
13  ' 


Find  the  greatest  and  least  pulls  upon 


one  of  the  chains.  The  ends  of  the  chains  are  attached  to  saddles  on 
rollers  on  the  top  of  piers  50  ft.  higli,  and  the  bacl< -stays  are  anchored 
50  ft.  from  the  foot  of  each  pier.  Find  the  load  upon  the  pier  j  and  the 
pull  upon  the  anchorage. 

^bis.  255  tons  ;  243I  tons  ;  637^  tons  ;  344.6  tons. 

20.  A  bridge  444  ft.  long  consists  of  a  central  span  of  180  ft.  and  two 
side  spans  each  of  132  ft.  ;  each  side  of  the  platform  is  sus|)ended  by 
vertical  rods  from  two  iron-wire  cables;  each  pair  of  cables  passes  over 
two  masonry  abutments  and  two  piers,  the  former  being  24  ft.  and  the 
latter  39  ft.  above  the  surface  of  the  groand  ;  the  lowest  point  of  the 
cables  in  each  span  is  19  ft.  above  the  ground  surface  ;  at  the  abutments 
the  cables  are  connected  with  straight  wrought-iron  chains,  by  means  of 
which  they  arc  attached  to  anchorages  at  a  horizontal  distance  of  66  ft. 
from  the  foot  of  each  abutment  ;  the  dead  weight  of  the  bridge  is  3500 
lbs.  per  lineal  foot,  and  the  bridge  is  covered  with  a  proof  load  of  4500 
lbs.  per  lineal  foot.     Determine — 

{a)  The  stresses  in  the  cables  at  the  points  of  support  and  at  the 
lowest  points. 

{b)  The  dimensions  and  weights  of  the  cables  (i)  if  of  uniform  sec- 


I  if 


738 


THEORY  OF  STRUCTURES. 


tion  throughout ;  (2)  if  each  section  is  proportioned  to  the  pull  across  it. 
(Unit  stress  =  14.958  lbs.  per  square  inch.) 

(<r)  The  alteration  in  the  length  of  the  tables  and  the  corresponding 
depression  of  the  platform  at  the  centre  of  each  span,  due  (i)  to  a  change 
of  60"  F.  from  llie  mean  temperature  ;  (2)  to  the  total  load,  E  being 
30,000,000  lbs. 

{d)  The  pressure  and  bending  mome;it  at  the  foot  of  pier. 

{e)  The  mass  of  masonry  in  tlie  anchorage  necessary  to  resist  the 
tendency  to  overturning  and  to  horizontal  displacement. 

Data. — Weight  of  masonry  per  cubic  foot  =  128  lbs. ;  safe  compres- 
sive stress  per  square  foot  —  12,000  lbs.  ;  coefficient  of  friction  =76; 
deviation  of  centre  of  pressure  in  base  of  pier  from  centre  of  figure  =  \ 
X  thickness  of  base. 


22  1 1 

Ans. — (a)  Side  span,  T\—r.^=i  =  //,  =  387,200  lbs.  =  T-,- , —  , 

V509  V146 


r. 


and  Ti  being  the  tensions  in  a  cable  at  summits  of 

low  and  high  piers,  respectively  ;  centre  span,  T-~ 

V'97 
=  405,000  lbs.  1=  H,  T  being  tension  at  summit  of  high 
pier. 
(J))  Side  span  :  Length  =  1352*2  f^-  '<  sect,  area  at  summit 
of  high  pier  =  28.43  sq.  in.  ;  weight  if  of  uniform  section 
=  12,834.  lbs.,  if  proportioned  to  pull  ~  11,710  lbs. 
Centre  span  :  Length  =  iSjff  ft. ;  sect,  area  at  summit 
of  pier  =  29.6  sq.  in.  ;  weight  if  of  uniform-  section 
=  18,361  lbs.,  if  proportioned  to  pull  =  17,267  lbs. 
(c)  (1)  .0594  ft.  for  side  span  and  .0775  ft.  for  centre  span  ; 

(2)  .0675 '       "     .0927 

(</)  High  pier:    Overturning    moment  =  694,200  ft. -lbs. ; 
bearing    area    at    summit  =  iiSf    sq.  ft.;   thickness 
=  8  ft.;    uniform  width  =  I4f  ft.;   thickness  of  base 
=  10.7  ft.  ;  weight  of  pier  =  692,348.8  lbs.  ;  total  pres- 
sure on  base  =  2,1 16,348.8  lbs. 
{e)  Weight  to  resist   upward  pull  =  29,333^  lbs. ;  weight 
to  resist  horizontal  displacement  =  509,474  lbs. 
21.  In  the  preceding  question,  if  the  piers  are  wrought-iron  oscillat- 
ing columns,  and  if  equilibrium,  under  an  unequally  distributed  load.  i< 
maintained  by  connecting  the  heads  of  the  columns  with  each  other  ami 
with  tlic  abutments  by  iron-wire  stays,  determine  the  proper  dimensions 
of  the  stays,  assuming  them  approximately  straight.     Assume  that  the 
proof  load  covers  (a)  a  side  span  ;  {/>)  two  side  spans ;  (c)  the  centre  span. 
Ans. — {a)  Pull  on  stays  in  centre  span  =  840.050  lbs. 

(^h)      =  double  that  in  (a). 

(c)      "      ' sids  span      —  948,432  lbs. 


EXAMPLES. 


739 


22.  A  floating  landing-stage  is  held  in  position  by  a  number  of  4|- 
lii.  steel-wire  cables  anchored  to  the  shore,  a  shoreward  movement  being 
prevented  by  rigid  iron  booms,  pivoted  at  the  ends  and  stretching  from 
sh(<re  to  stage.  The  difference  of  level  between  the  shore  and  stage  at- 
tachments of  the  cables  is  50  ft.,  and  the  horizontal  distance  between 
these  points  is  150  ft.  The  horizontal  pull  upon  each  cable  is  1360  lbs. 
Find  the  length  of  the  cable,  and  the  tensions  at  the  points  of  attach- 
ment. (Weight  of  cable  =  490  lbs.  per  cubic  foot ;  form  of  cable  a  com- 
mon catenary.)  Ans.  342.82  ft.;  12,267.2  lbs.  and  10,132  lbs. 


!>\ 

f  1 

1 

" 

,'\ 

11 

\ 

! 

\ 

1 

"\ 

r 

:N!    M 


CHAPTER   XIII. 


ARCHES   AND   ARCHED    RIBS. 


I.  An   arch    may  be   constructed  of   masonry,  brickwork, 
timber,  or  metal. 


Fig.  478. 

In  the  figure  ABCD  represents  the  profile  of  an  arch.  The 
under  surface  AD  is  called  the  soffit  or  intrados.  The  upper 
surface  BC  is  sometimes  improperly  called  the  cxtrados.  The 
highest  point  K  o{  the  soffit  is  the  croton  or  key  of  the  arch. 
The  spriugings  or  skeivbacks  are  the  surfaces  AB^  DC  from 
which  the  arch  springs,  and  the  haunches  are  the  portions  of 
the  arch  half-way  between  the  springings  and  the  crown. 
Upon  each  of  the  arch  faces  stands  a  spandril  wall,  and  the 
space  between  these  two  external  spandrils  may  be  occupied 
by  a  series  of  internal  spandrils  spaced  at  definite  distances 
apart,  or  may  be  filled  up  to  a  certain  level  with  masonry  (i.e.. 
backing)  and  above  that  with  ordinary  ballast  or  other  rough 
m.aterial  (\.e.,jiiling). 

A  masonry  arch  consists  of  courses  of  wedge-shaped  blocks 
with  the  bed-joints  perpendicular,  or  nearly  so,  to  the  soffit. 

740 


EQUll.lHKATED   POl.YGOX  AXD   I.IXE   OF  RESISTAXCE.     74^ 

Tlic  blocks  arc  called  voussoirs,  and  the  voussoirs  at  tli'  crown 
are  the  keystones  of  the  arch. 

A  Prick  arch  is  usually  built  in  a  number  of  rings. 

Consider  the  portion  of  the  arch  houndetl  by  the  vertical 
I)lane  KE  at  the  key  and  by  tlie  plane  .//>'. 

It  is  kept  in  equilibrium  by  the  reaction  R  at  KE,  the  reac- 
tion A^,  at  Ali,  .Mid  the  wei^dit  K,  of  the  portion  under  con- 
sideration and  Its  superincumbent  load. 

Let  ^'  and  T  be  the  points  of  application  of  A*,  and  R, 
respectivel}'. 

Let  the  directions  of  A,  and  A  intersect  in  a  point.  The 
direction  of  F,  must  also  pass  through  the  same  point. 

Taking  moments  about  S, 

/>,  and  J',  being  the  perpendicular  distances  of  the  directions  of 
A  and   \\  from  S,  respectively. 

Similarly,  the  portion  KECD  of  the  arch  gives  the  equation 

F,  being  the  weight  to  which  it  is  subjected,  and  p^,  )\  the 
perpendicular  distances  of  the  directions  of  A  and  K,  from  the 
point  of  application  K  of  the  reaction  at  the  plane  DC. 

If  the  arch  and  the  loading  are  symmetrical  with  respect 
to  the  plane  KE, 

^1  =  ^!i'    Ji  =J's»     ^n^  therefore  /,  =■  p^. 

Hence  the  direction  of  A  will  be  horizontal,  which  might  have 
been  inferred  by  reason  (^f  the  symmetry. 

The  magnitudes  of  the  reactions  are  indeterminate,  as  the 
positions  of  the  pnintr.  of  application  {S,  T,  V )  are  arbitrary, 
and  can  only  be  fixed  by  a  knowledge  of  the  law  of  the  varia- 
tion of  the  stress  in  the  material  at  the  bounding  planes  AB, 
KE. 

2.  Equilibrated  Polygon  and  Line  of  Resistance. — 
Suppose  an  arch  divided  into  a  number  of  elementar}-  portions 


r-- 


n 


74^ 


rilEOKY   Oh    STRUCTURES. 


kc' ,  k'e"  .  .  ,  Cc.g.,  tlic  vousaoirs  of  a  masonry  arch)  by  a  scries 
of  ioints  kc,  k'c'  .  .  . 


Fig.  479. 


Fig.  480. 


Let  Wt,  IVj,  ...  be  the  loads  directly  supported  by  the 
several  portions.  These  loads  generally  consi.st  of  the  wciijlu 
of  a  portion  (e.g.,  ke')  -f-  the  weiglit  of  the  superiiicumbciit 
mass  +  the  load  upon  the  overlying  roadway  ;  the  lines  of 
action  of  the  loads  are,  therefore,  nearly  always  vertical. 

Each  elementary  portion  may  be  considered  as  acted  upon 
and  kept  in  equilibrium  by  t/inr  forces,  viz.,  the  external  load 
and  the  pressures  at  the  joints.  If  the  pressure  and  its  point 
of  application  at  any  given  joint  have  been  determined,  the 
pressures  and  the  corresponding  points  of  application  at  the 
other  joints  may  also  be  found. 

For,  let  I  2  34  .  .  .  be  the  line  of  loads,  so  that  12=  W,, 

2  3  =  n\ ,  .  .  . 

Assume  that  the  pressure  P  and  its  point  of  application  r 
at  any  given  joint  ke  are  known. 

Draw  o  i  to  represent  P  in  direction  and  magnitude. 

Then  02  evidently  represents  the  resultant  of  /^and  W,  in 
direction  and  magnitude,  and  this  resultant  must  be  equal  and 
opposite  to  the  pressure  /•,  at  the  joint  k'/. 

Hence,  a  line  n'n  drawn  through  «,  tlie  intersection  of  P 
and  W,,  parallel  to  20,  is  the  direction  of  the  pressure  P,,  and 
intersects  k'e'  in  the  point  of  application  r'  of  P, 

Again,  o  3  represents  the  resultant  of  /'  and  IV^  in  direc- 
tion and  magnitude,  and  this  resultant  must  be  equal  and 
opposite  to  the  pressure  /',  at  the  joint  k"e". 

The  line  «"«'  drawn  through  ;/',  the  intersection  of  P^  and 


EQUILIBRATED  POLYGOX  AND  LINE   OF  A'ES/.S lAXCE.     743 


VV,„  parallel  to  30,  is  the  direction  of  the  pressure  /',  and  inter- 
sects k"i'"  in  the  point  of  application  r"  of  J\. 

Proceeding  in  this  manner,  a  scries  of  points  of  application 
or  ccntris  of  resistance  r',  r",  r"\  .  .  .  may  be  found,  the 
corresponding  pressures  being  represented  by  02,  03,04,  .  .  . 

T\\c  polygon  of  pressures  formed  by  the  lines  of  action  of 
P,Pi,  F,,  .  ,  .  \s  termed  an  equilibrated poly<j;on,  and  is  a  funic- 
ular polygon  of  the  loads  upon  the  several  portions. 

The  polygon  formed  by  jcining  the  points  r,  r' ,  r",  .  .  . 
successively,  is  called  the  line  of  resistance. 

In  the  limit,  when  the  joints  are  supposed  indof^n'tely  near, 
these  polygons  become  curves,  the  curve  in  the  d;  of  the 
equilibrated  polygon  being  known  as  a  linear  arch. 

The  two  curves  ma}',  without  sensible  erro  be  s'lpposeH 
identical,  and  they  will  exactly  coincide  if  the  joints  (cf  cf  use 
imaginary  it^  such  a  case)  are  made  parallel  to  the  lines  of 
action  of  the  external  loads.  This  may  be  easily  proved  as 
foh'./s: 

Let  the  figure  represent  a  portion  of  an  arch  bounded  by 
th.e  joints  (imaginarj-)  KE,  JILV  paralh.-l  to  the  lines  of  action 
of  the  external  loads,  which  will  be  assumed  vertical. 

Reduce  the  superincumbent  loads  to  an  equivalent  mass  of 
arch  material. 

Let  //,  e.g.,  be  the  depth  of  material  of  specific  weight  w, , 
overlying  the  arch  at  any  given 
point,  and  let  Q  be  the  load  per 
unit  of  area  of  roadway. 

Also,  let  la  be  the  specific 
weight  of  the  arch  material. 

Then  x,  the  equivalent  depth, 
is  given  by 


Fig.  4G1. 


WX  =  XV  Jl  -\-  Q. 

If  the  value  of  x  is  deter- 
mined at  different  points  along  the  arch,  a  \i\c\u\\,  en  maybe 
obtained  defining  a  mass  ENne  of  arch  material  \vhich  may  be 
substituted  for  the  superincumbent  load.  Denote  the  weight 
of  the  mass  MKen  by  If. 


744 


THEORY  OF  STRUCTURES. 


Let  the  pressure  P  and  its  point  of  application  0  at  tlie 
joint  KE  be  given. 

Take  O  as  the  origin,  the  Hne  OA  in  the  direction  of  Pas 
the  axis  of  x,  and  the  vertical  through  0  as  the  axis  of  y,  and 
let  ^  be  the  angle  between  the  two  axes. 

Let  the  lines  of  action  of  /*  and  IF  intersect  in  G.  The 
line  of  action  of  their  resultant  will  intersect  MN  m  the  centre 
of  resistance  O^ 

Let  X,  I'' be  the  co-ordinates  of  O^. 

Let  z  be  the  depth  of  an  elementary  slice  of  thickness  dx, 
parallel  to  OK  at  any  abscissa  x.  Its  weight  =  ivzdx  sin  f^. 
Then 

W.OG  =  fwzdx  s[nd.x=  W{X -  A G). 

P      AG      AG 
But  -,Ty.=  -7—,  =  — TT-,  since  the  triangle  AGO  is  evidently  a 

triangle  of  forces  for  the  forces  acting  upon  the  mass  under 
consideration. 
Also, 

/A- 
zv2dx  .  sin  6^. 

.-.    rivsxdx  .  sin  ^  ==  IVX  -  W~Y=  X  Cws  sin  edx  -  PY. 

This  is  the  equation  to  the  line  of  resistance. 
Taking  the  differential  of  this  equation, 

ws'Xsm  OdX  =  Xu's'  sin  ft/A'+  IVdX  -  PdY, 

z'  being  the  depth  corresponding  to  the  abscissa  X. 

dY        IV       AO,       ^        .„^ 
...-=-  =  -^-tan^^O.. 

Thus  the  tangents  to  the  curve  of  pressures  and  to  the 
curve  of  centres  of  pressure  at  any  given  point  coincide,  and 
the  curves  must  therefore  also  coincide. 


3.  Con 

a  portion  o 
joints  (real 
Let  W 
bent  load, 
the  reactio; 
intersect  M 
tion  of  IV '\ 
the  reactio  I 
must  also  p; 
direction  in 
order  that  t 
stable,  three 
fulfilled,  viz. 
First.    Ti 
may  be  no  t 
Second.  " 
the  angle  be 
must  not  ex( 
the  arch  is  c( 
N.B.~T\ 
30°. 

Third.  T, 
must  not  exec 
Further,  t 
case  of  mas 01 
at  every  poin 
tensile  forces, 

The  best 

as  the  pressu 

area  PQ.     It 

tribution,  and 

varies  uniforn 

With  this  . 

Let /be  t 

the  most  com 

Let   OS^ 

coefficient  wh 


CONDITIONS  OF  EQUILIBRIUM. 


743 


3.  Conditions  of  Equilibrium. — Let  the  figure  represent 
a  portion  of  an  arch  of  thickness  unity,  between  any  two  bed- 
joints  {real  or  imaginary)  MiV,  PQ. 

Let  W^be  its  weight  together  with  that  of  the  superincum- 
bent load.  Let  the  direction  of 
the  reaction  R'  at  the  joint  AIN 
intersect  MN  in  m  and  the  direc- 
tion of  W '\w  H.  For  equihbrium, 
the  reaction  R"  at  the  joint  PQ  1\ 


must  also  pass  through  n.    Let  its     ^\ 
direction  intersect  PQ  in   O.     In 
order  that  the  equilibrium  may  be 
stable,  three   conditions  must   be 
fulfilled,  viz. : 

first.  The  point  O  vinst  lie  between  P  and  Q,  so  that  there 
may  be  no  tendency  to  turn  about  the  edges  P  a.nd  Q. 

Second.  There  must  be  no  sliding  along  PQ,  and  therefore 
the  angle  between  the  direction  of  R"  and  the  normal  to  PQ 
must  not  exceed  the  angle  of  friction  of  the  material  of  which 
the  arch  is  composed. 

N.B. — The  angle  of  friction  for  stone  upon  stone  is  about 

30°. 

Third.  The  maxiimun  intensity  of  stress  at  any  paint  in  PQ 
must  not  exceed  the  safe  resistance  of  the  material. 

Further,  the  stress  should  not  change  in  character,  in  the 
case  of  masonry  and  brick  arches,  but  should  be  a  compression 
at  every  point,  is  these  materials  are  not  suited  to  withstand 
tensile  forces. 

The  best  position  for  O  would  be  the  middle  point  of  PQ. 
as  the  pressure  would  then  be  uniformly  distributed  uver  the 
area  PQ.  It  is,  however,  impracticable  to  insure  such  a  dis- 
tribution, and  it  has  been  sometimes  assumed  that  the  stress 
varies  uniformly, 

With  this  assumption,  let  A'^  be  the  normal  component  of  R''. 

Let  /  be  the  maximum  compressive  stress,  i.e.,  the  stress  at 
the  most  compressed  edge,  e.g.,  P. 

L,et  OS  —  q  .  PQ,  S  being  the  middle  point  of  PQ,  and  q  a 
coefflcient  whose  value  is  to  be  determined. 


p    ■: 


H' 

:.  .1 

j  J 

i   1 


t 


746 


THEORY  OF  STRUCTURES. 


Then  \{  PO  <—, 
3 


N  =  y,PO^\f.PQ{\-qY 


MPO>^A 
3 


f.PQ 
1+6^' 


PQ 
and  in  the  limit  when  PO  =  — ,  i.e.,  when  the  intensity  of 

stress  varies  uniformly  from /at  Pto  «//at  Q, 


1 


and     iV: 


/•PQ 


(See  Art.  i6,  Chap.  IV.) 

Similarly,  if  Q  is  the  most  compressed  edge,  the  limiting 
position  of  O,  the  centre  of  resistance  or  pressure,  is  at  a  point 

PQ 
a  defined  b>  QO'  =  -^. 

Hence,  as  there  should  be  no  tendency  on  the  part  of  the 

joints  to  open  at  either  edge,  it  is  inferred   that  PO  or  QO' 

PO 
should  be  >  — -,  i.e.,  that  the  point  O  should  lie  within  the 

middle  third  of  the  joint. 

Experience,  however,  shows  that  the  "  middle-third " 
theory  cannot  be  accepted  as  a  solution  of  the  problem  of 
arch  stability,  and  that  its  chief  use  is  to  indicate  the  proper 
dimensions  of  the  abutments.  Joint  cracks  are  to  be  found  in 
more  than  90$^  of  the  arches  actually  constructed,  and  cases 
may  be  instanced  in  which  the  joints  have  opened  so  widely 
that  the  whole  of  the  thrust  is  transmitted  through  the  cds^es. 
In  Telford's  masonry  arch  over  the  Severn,  of  150  ft.  span, 
Baker  discovered  that  there  had  been  a  settlement  (15  in.) 
sufficient  to  induce  a  slight  reverse  curvature  at  the  crown  of 
the  sofifit.  Again,  the  position  of  the  centre  of  pressure  at  a 
joint  is  indeterminate,  and  it  is  therefore  impossible  as  welt  as 
useless  to  make  any  calculations  as  to  the  maximum  intensity 
of  stress  due  to  the  pressure  at    the   joint.     What   seems  to 


happen  1 
exceeds 
formly  di 
curve  of 
arch  are 
been  dec 
Baker  ma 

Let    : 
width  of 

Let/ 
square  foe 

An  arc 

surfaces  a 

sures,  can 
advance  w 
were  possi 
compressil 
ments,  ho 
reliance  ca 
values. 

4.  Joini 
between  w] 


the  centre 
pressure  m; 
different  va! 

Let  ^y 
horizontal  t 
X  is  called  t 

The   an{ 


i 


if  I 


JOINT  OF  RUPTURE. 


747 


happen  in  practice  is,  that  the  straining  at  the  joints  generally 
exceeds  the  limit  of  elasticity,  and  that  the  pressure  is  uni- 
formly distributed  for  a  certain  distance  on  each  side  of  the 
curve  of  pressures.  Thus,  the  proper  dimensions  of  a  stable 
arch  are  usually  determined  by  empirical  rules  which  have 
been  deduced  as  the  results  of  experience.  For  example, 
Baker  makes  the  following  statement : 

Let  T  be  the  thrust  in  tons  or  pounds  per  lineal  foot  of 
width  of  arch. 

Let  /  be  the  safe  working  stress  in  tons  or  pounds  per 
square  foot. 

An  arch  will  be  stable  if  an  ideal  arch,  with  its  bounding 

surfaces  at  a  minimum  distance  of  —  —  from  the  curve  of  pres- 

sures,  can  be  traced  so  as  to  lie  within  the  actual  arch.  An 
advance  would  be  made  towards  a  more  correct  theory  if  it 
were  possible  to  introduce  into  the  question,  the  elasticity  and 
compressibility  of  the  materials  of  construction.  These  ele- 
ments, however,  vary  between  such  wide  limits  that  no 
reliance  can  be  placed  upon  the  stresses  derivable  from  their 
values. 

4.  Joint  of  Rupture. — Let  i  2,  3  4  be  the  bounding  surfaces 
between  which  the  curve  of  pressures  must  lie,  and  let  4  be 


Fig.  483. 

the  centre  of  pressure  at  the  crown.  A  series  of  curves  of 
pressure  may  be  drawn  for  the  same  given  load,  but  with 
different  values  of  the  horizontal  thrust  h. 

Let  ^^y  be  that  particular  curve  which  for  a  value  H o[  the 
horizontal  thrust  is  tangent  to  the  surface  I  2  at  4r ;  the  joint  at 
X  is  called  \.\\^  joint  of  rupture. 

The   angle   which   the   joint  of   rupture   mgkes  with   the 


i;ii^! 


m.» 


l!: 


748 


THEORY  OF  STRUCTURES. 


horizontal  is  about  30°  in  semicircular  and  45°  in  elliptic 
arches. 

The  position  of  the  joint  in  any  given  arch  may  be  tenta- 
tively found  as  follows : 

Let  J  be  any  joint  in  the  surface  i  2. 

Let  ^Fbe  the  weight  upon  the  arch  between  y  and  i. 

Let  X  be  the  horizontal  distance  between  J  and  the  centre 
of  gravity  of  W. 

Let  y  be  the  vertical  distance  between  J  and  4. 

It  will  also  be  assumed  that  the  thrust  at  4  is  horizontal. 

If  the  curve  of  pressure  be  now  supposed  to  pass  through 
y,  the  corresponding  value  of  the  horizontal  thrust  Ji  is  given 
by 

hY  =z  WX. 

By  means  of  this  equation,  values  of  h  may  be  calculated 
for  a  number  of  joints  in  the  neighborhood  of  the  haunch,  and 
the  greatest  of  these  values  will  be  the  horizontal  thrust  H  for 
the  joint  x.  This  is  evident,  as  the  curve  of  pressure  for  a 
smaller  value  of  k  must  necessarily  fall  bcloiv  4x_y. 

When  this  happens,  the  joints  will  tend  to  open  at  the 
lower  edge  of  the  joint  I  4  and  at  the  upper  edges  of  the  joints 
at  .V  and  at  2  3,  so  that  the  arch  may  sink  at  the  crown  and 
spread,  unless  the  abutments  and  the  lower  portions  of  the 
arch  are  massive  enough  to  counteract  this  tendency. 

If  the  curve  of  pressure  fall  aiove  ^xy,  an  amount  of  back- 
ing sufificient  to  transmit  the  thrust  to  the  abutments  must  be 
provided.  The  same  result  may  be  attained  by  a  uniform  in- 
crease in  the  thickness  of  the  arch  ring,  or  by  a  gradual  increase 
from  the  crown  to  the  abutments. 

For  example,  the  upper  sur- 
face (extrados)  of  the  ring  for  an 
arch  with  a  semicircular  soffit 
A  KB,  having  its  centre  at  O,  may 
be  delineated  in  the  following 
manner: 

Let  X  define  the  joint  of  rup- 
ture in  the  soffit ;     then  AOx  =  30°. 


In  Ox 

ness  at  the 
The  an 

duced  may 

ring,  and  tl 

xf  to  the  ai 
5.  Mini 

resultant 

rectangular 
Let  y  b 

tion  from 
Let  //  a 

components 
Let  zv  b 

in  the  abuti 
Let  h  be 
Let  /  be 
In  order 

toe  D,  the 

to  D  ph 

than  the  mo 


the  n 

lus  thi 


This  relai 
abutment  mj 


which  define; 
ment. 


MINIMUM    THICKNESS  OF  ABUTMENT. 


749 


In  Ox  produced  take  xx'  =  2  X  KD,  KD  being  the  thick- 
ness at  the  crown. 

The  arc  Dx'  of  a  circle  struck  from  a  centre  in  DO  pro- 
duced may  be  taken  as  a  part  of  the  upper  boundary  of  the 
ring,  and  the  remainder  may  be  completed  by  the  tangent  at 
x!  to  the  arc  Dx' . 

5.  Minimum  Thickness  of  Abutment. — Let    T  be   the 
resultant  thrust  at  the  horizontal  joint  BC  of  a 
rectangular  abutment  A  BCD. 

Let  y  be  the  distance  of  its  point  of  applica- 
tion from  B. 

Let  //  and  V  be  the  horizontal  and  vertical 
components  of  T. 

Let  w  be  the  specific  weight  of  the  material 
in  the  abutment. 

Let  Ji  be  the  height  AB  of  the  abutment. 

Let  t  be  the  width  AD  of  the  abutment.  F'c  485. 

In  order  that  there  may  be  no  tendency  to  turn  about  the 
toe  D,  the  moment  of  the  weight  of  the  abutment  with  respect 
to  D  plus  the  moment  of  V  with  respect  to  D  must  be  greater 
than  the  moment  of  H  with  respect  to  D.     Or, 


wht-^  V{t-y)>  Hk, 


■1 

i'  1 

%:\ 

^  1 

m-i 


lit 


lllli!5 


or 


/  + 


z^V 


2//  ,  2  F      ,     F' 

— -r  - —  y  -\  —. — • 
zv       wh  zv'h* 


This  relation   must  hold  good  whatever  the  height  of  the 
abutment  may  be ;  and  if  h  is  made  equal  to  00 , 


t> 


V     w 


which  defines  a  minimum  limit  for  the  thickness  of  the  abut- 
ment. 


750 


THEORY  OF  STRUCTURES. 


6.  Empirical  Formulae.— In  practice  the  thickness  t  at 
the  crown  is  often  found  in  terms  of  s,  the  span,  or  in  terms  of 
p,  the  radius  of  curvature  at  the  crown,  from  the  formulae 

t  ■=.  c  V'j,     or     /  =  Vcp, 
t,  s,  and  p  being  all  in  feet,  and  c  being  a  constant. 

According  to  Dupuit,  i  =  .36  ^s  for  a  full  arch ; 

/  =  .27  i^s  for  a  segmental  arch. 
According  to  Rankine,    =  i^.i  2/3  for  a  single  arch  ; 

/  =  V.I 7/3  for  an  arch  of  a  series. 

7.  Examples  of  Linear  Arches,  or  Curves  of  Pressure. 
(a)  Linear  Arch  in  the  Form  of  a  Parabola, — Suppose  that 

the  cable  in  Art.  4,  Chap.  XII,  Case  B,  is  exactly  inverted, 
and  that  it  is  stiffened  in  such  a  manner  as  to  resist  distortion. 
Suppose  also  that  the  load  still  remains  a  uniformly  distributed 
weight  of  intensity  w  per  horizontal  unit  of  length  A  thrust 
will  now  be  developed  at  every  point  of  the  inverted  cable 
equal  to  the  tension  at  the  corresponding  point  of  the  original 
cable.  Thus  the  inverted  parabola  is  a  linear  arch  suitable  for 
a  real  arch  which  has  to  support  a  load  of  intensity  zo  per 
horizontal  unit  of  length. 

The  horizontal  thrust  at  the  crown  ■=.  H  =■  wp, 

p  being  the  radius  of  curvature  at  the  crown. 

{l>)  Linear  Arch  in  the  Form  of  a  Catenary.     Transformed 

Catenary. — If  the  cable  in  Art.  4,  Chap.  XU,  Case  A,  is  in- 

1^  p      Q  K,      verted  and  stiffened  as  before,  a  linear 

arch  is  obtained  suitable  for  a  real 
arch  which  has  to  support  a  load  dis- 
tributed in  such  a  manner  that  tiie 
weight  upon  any  portion  AP  is  pro- 
{  ,  portional  to  the  length  of  AP,  and  is 
in  fact  =/'^.  The  area  OAPN  =^  ins. 
Thus,  a  lamina  of  thickness  unity 
and  specific  weight  zv,  bounded  by  the  curve  AP,  the  directrix 
ON,  and  the  verticals  AO,  PN,  weighs  xojns,  and  may  be  taken 


Fig.  486. 


to  repre; 
i.e.,  if  th 

Th( 

the  radii 
A  dis 
catenary 
through 
sary  that 
meet  the 
may  be  o 
Upon 
horizonta 
generatin 
Cut  tl 
quired  an, 
solid  will 
arch,  and 
the  arch  v 
lamina  art 
The  pr 
length. 

The  pi 

of  the  line; 

of  the  ang! 

Let  X, 

formed  cat, 

Let  X,  _ 

in  the  cate 

Let  A'( 

Then 


he  equp.tii 


EXAMPLES  OF  LINEAR  ARCHES. 


751 


lis 
\s. 

y 

Ix 
In 


to  represent  the  load  upon  the  arch  if  %vms  =  ps,  i.e.,  if  zvm  —P, 
i.e.,  if  the  weight  of  m  units  of  the  lamina  is  w. 

The  horizontal  thrust  at  the  crown  =  //  =  wvi  =■  zcp, 

the  radius  of  curvature  (p)  at  the  crown  being  equal  to  ///. 

A  disadvantage  attached  to  a  linear  arch  in  the  form  of  a 
catenary  lies  in  the  fact  that  only  o/ie  catenary  can  pass 
through  tzvo  given  points,  while,  in  practice,  it  is  often  neces- 
sary that  an  arch  sliall  pass  through  ^/ine  points  in  order  to 
meet  the  requirements  of  a  given  rise  and  span.  This  difificulty 
may  be  obviated  by  the  use  of  the  transfornied  catenary. 

Upon  the  lamina  PAPNN  as  base,  erect  a  solid,  with  its 
horizontal  sections  all  the  same,  and,  for  simplicity,  with  its 
generating  line  perpendicular  to  the  base. 

Cut  this  solid  by  a  plane  through  A'N  inclined  at  any  re- 
quired angle  to  the  base.  The  intersection  of  the  plane  and 
solid  will  define  a  transformed  catenary /''.<4'/'',  or  a  new  linear 
arch,  and  the  shape  of  a  new  lamina  P'A'P'NN,  under  which 
the  arch  will  be  balanced.  This  is  evident,  as  the  new  arch  and 
lamina  are  merely  parallel  projections  of  the  original. 

The  projections  of  horizontal  lines  will  remain  the  same  in 
length. 

The  projections  of  vertical  lines  will  be  c  times  the  lengths 
of  the  lines  from  which  they  are  projected,  c  being  the  secant 
of  the  angle  made  by  the  cutting  plane  with  the  base. 

Let  X,  F  be  the  co-ordinates  of  any  point  /''  of  the  trans- 
formed catenary. 

Let  X,  y  be  the  co-ordinates  of  the  corresponding  point  P 
in  the  catenary  proper. 

Then 

y_P^_     _A'0  _M 

y~  FN  ~^~  A0~'^« ^^' 

The  equation  to  the  catenary  proper  is 


m 


% 


■P 


lit 


2  \ 


(2) 


752  THEORY  OF  STRUCTURES. 

Substituting  in  the  last  equation  the  value  oi  y  given  by  eq.  (i), 


and  her 


\ 


(3) 


which  is  the  equation  to  the  transformed  catenary. 

With  this  form  of  linear  arch  the  depths  M  over  the  crown 
and  Fover  the  springings,  for  a  span  2x%  may  be  assumed,  and 
tlic  corresponding  value  of  m  determined  from  eq.  (3). 

It  is  convenient,  in  calculating  ;«,  to  write  eq.  (3)  in  the  form 


X 

m 


=  iog.|j^+y^,-i|.  .  .  . 


(4) 


The  slope  ?'  at  /"  is  given  by 


tan  i' 


.,      dV      Ml-        -'-\      Ms 
ax      2)fn  I        m 


s  being  the  length  AP  of  the  catenary  proper,  corresponding 
to  the  length  A'P'  of  the  transformed  catenary. 


The  area  OA 


'p'N=  r 


'■*,,,         Mm 

Ydx  =  - 


\e"'  —  e    '") 


Ms. 


The  triangle  P'  TN  is  a  triangle  of  forces  for  the  portion 
A'P'. 

Tlie  triangle  PTN  is  a  triangle  of  forces  for  the  portion  AP. 

(The  tangents  at  7^  and  P'  must  evidently  intersect  ON  in 
the  same  point  T.) 

Let  H'  be  the  horizontal  thrust  at  A',  //being  that  at  A. 

Let  P'  be  the  weight  upon  A'P',  /'being  that  upon  AP. 

Let  R'  be  the  thrust  ai  P'. 

Then 

P'  _  area  OA'P'N _Ms  _  M 
•     ■         ~P  ~  area  OAPN  ~~  nis  ~~  m  ' 


Thei 


and  the  r 

of  the  tra 

The  t 

to  a  linea 

(c)  Ch 

which  has 

normal  pn 

sity  shouk 

Consid 

element    L 

■turned      t( 

straight. 

Let  the 
ant  pressur 
Let  CE 
CD. 

The  an^ 
The  hor 
This  is  < 

•'.  the  he 

Similarh 

pressure  =^ 


and  hence 


EXAMPLES  OF  LINEAR  ARCHES. 


^        M  M  .^ 

t^  =  — /^=  —wms  =  wMs ; 
tn         m 


753 


y 


A 


in' 


H  =  F  cot  i'  =  wMs-^j-  =  wne  =  H\ 

Ms 


R'=  H'  sec  i'  =wrn'\/  i  + 


/; 


w  Vm*  +  APs' . 


m 
The  radius  of  curvature  p'  at  the  crown  =  ir=^ 


M' 


H'  =  wMp'  =  H—  wp, 


and  the  radius  of  the  "  catenary  proper  "  is  M  times  the  radius 
of  the  transformed  catenary. 

The  term  "  equiHbrated  arch  "  has  generally  been  applied 
to  a  linear  arch  with  a  horizontal  extrados. 

{c)  Circular  and  Elliptic    Linear  Arches. — A   linear   arch 
which  has  to  support  an  external 
normal  pressure  of  uniform  inten- 
sity should  be  circular. 

Consider  an  indefinitely  small 
element  CD,  which  meiy  be  as- 
sumed to  be  approximately 
straight. 

Let  the  direction  of  the  result- 
ant pressure  upon  CD,  viz.,  / .  CD,  make  an  angle  6  with  OB. 

Let  CE,  DE  be  the  vertical  and  horizontal  projections  of 
CD. 

The  angle  DCE  =  ft. 

The  horizontal  component  of />.  CD  =/.  CD  cos  0  =p.  CE, 

This  is  distributed  over  the  vertical  projection  CE. 

.'.  the  horizontal  intensity  of  pressure  =  p  .  CE  -^  CE  ■==■  p. 

Similarly,  it  may  be  shown  that  the  vertical  intensity  of 
pressure  =/>. 


Fig.  487. 


li' 

w 

1 

1 

\ 

1 

\  ['.' 

\ 

i 

i 

'lilllliil 


754  THEORY  OF  STRUCTURES. 

Thus,  at  any  point  of  the  arch, 

the  horizontal  intensity  of  pressure 

=  vertical  intensity  =  normal  intensity  =/. 
Again,  the  total  horizontal  pressure  on  one-half  of  the  arch 

=  2(/ .  CE)  =  p2{CE)  =pr  =  H, 
and       the  total  vertical  pressure  on  one-half  of  the  arch 
=  2(/> .  DE)  =  p2{DE)  =ipr  =  P. 

Hence,  at  any  point  of  the  arch  the  tangential  thrust  =  pr. 

Next,   upon  the  semicircle  as  base,  erect  a  semi-cylinder. 
Cut  the  latter  by  an  inclined  plane  drawn  through  a  line  in  the 


Fig.  488. 

plane  of  the  base  parallel  to  OA,  The  intersection  of  the  cut- 
ting plane  and  the  semi-cylinder  is  the  semi-ellipse  B'AB',  in 
which  the  vertical  lines  are  unchanged  in  length,  while  the 
lengths  of  the  horizontal  lines  are  c  times  the  lengths  of  tlie 
corresponding  lines  in  the  semicircle,  c  being  the  secant  of  the 
angle  made  by  the  cutting  plane  with  the  base.  A  semi- 
elliptic  arch  is  thus  obtained,  and  the  forces  to  which  it  is  sub- 
jected are  parallel  projections  of  the  forces  acting  upon  the 
semicircular  arch. 

These  new  forces  are  in  equilibrium  (see  Corollary). 
Let  P"  =  the  total  vertical  pressure  upon  one-half  of  the 
arch  ; 
H'  =  the  total  horizontal  pressure  upon  one-half  of  the 
arch  ; 


EXAMPLES  OF  LINEAR  ARCHES. 

F 

Py  —  vertical  intensity  of  pressure  =  yrrr, ; 

pj  =  horizontal  intensity  of  pressure  =  tt-j-. 

Then 

P'  =  P=I/  =  pr] 


755 


0) 


p^ 

B' 


pr      /_ 


^'  ~  0B'~ c,OB~~  cr~  c'      •     •    •     •    (2) 


H' =  cH=cP=:cP'\ (3) 

H'       c.H     cpr 

^'  -OA'-aA-T-'^ w 


Hence,  by  eq.  (3), 


H' 


OB' 


P'  ~^  "  oa'' 


or,  the  total  horizontal  and  vertical  thrusts  are  in  the  ratio  of 
the  axes  to  which  they  are  respectively  parallel,  and,  by  eqs. 
(2)  and  (4), 


/  8    » 


OB 

or,  the  vertical  and  horizontal  intensities  of  pressure  are  in  the 
ratio  of  the  squares  of  the  axes  to  which  they  are  respectively 
parallel. 

Any  two  rectangular  axes  OG,  OK  in  the  circle  will  project 
into  a  pair  of  conjugate  radii  OG',  OK'  in  the  ellipse. 

Let  OG'  =zr„  OK'  =  r,; 

Q  =  total  thrust  along  elliptic  arch  at  K; 

Then 

H      r  ^     H      r 

"  R  ~r'  


I  I 

I 

Hi 


r 


l! 


:  •    ( 


m 


in  I 


756 


THEORY  OF  STRUCTURES. 


or,  the  total  thrusts  along  an  elliptic  arch  at  the  extremities  <a 
a  pair  of  conjugate  radii  are  in  the  ratio  of  the  radii  to  whicli 
they  are  respectively  parallel. 

The  preceding  results  show  that  an  elliptic  linear  arch  i> 
suitable  for  a  load  distributed  in  such  a  manner  that  the  vertical 
and  horizontal  intensities  (eqs.  (2)  and  (4))  at  any  point  of  tli' 
arch  are  unequal,  but  are  uniform  in  direction  and  magnitude. 

Corollary. — It  can  be  easily  shown  that  the  projected  forces 
acting  upon  the  elliptic  arch  are  in  equilibrium. 

The  equations  of  equilibrium  for  the  forces  acting  upon  the 
circular  arch  may  be  written 


«'(r|)+Frf.=o; 

7*  being  the  thrust  along  the  arch  at  the  point  xy,  and  X,  Y 
the  forces  acting  upon  the  arch  parallel  to  the  axes  of  x  and 
y,  respectively. 

If  T\  X\  V  be  the  corresponding  projected  forces, 

^  =  J,     Xds  =  cX'ds',     Yds  =  Y'ds'. 
as       as 

Hence,  the  above  equations  may  be  written 

d[^,cdx'^-\-cX'cis'  =0» 

d^-'dy)    +    FV/  =  o; 


and 


or 


and 


i-%) 


+    FV/=o. 


Hence,  the  forces  T\  X\  and  Y'  are  also  in  equilibrium. 


EXAMPLES  OF  LINEAR  ARCHES. 


757 


^<^o 


{d)  Hydrostatic  Arch. — Let  the  figure  represent  a  portion 
of  a  linear  arch  suited  to  support  a  load  ^  _ 
which  will  induce  in  it  a  normal  pressure  at 
every  point.  Tlie  pressure  beiii^  normal 
has  no  tangential  component,  and  tlie 
thrust  (7")  along  the  arch  must  therefore  be 
everywhere  the  same. 

Consider  any  indefinitely  small  element 
CD. 

It  is  kept  in  equilibrium  by  the  equal 
thrusts  (7")  at  the  extremities  C  and  D,  and  by  the  pressure 
p .  CD.  The  intensity  of  pressure  /  being  assumed  unk^m 
for  the  element  CD,  the  line  of  action  of  the  pressure /TcZ? 
bisects  C      i  right  angles. 

Let  the  normals  at  C  and  D  meet  in  (9, ,  the  centre  of 
curvature. 

Take  0,C  =  0,D  =  p,  and  the  angle  CO,D  =  2 JO. 

Resolving  along  the  bisector  of  the  angle  CO^D, 


Fig.  489. 


or 

and  hence, 


2Tsir\  ^6=  p.  CD  =  pp.  2 Ad, 
2TJ0=pp.2J0\ 

T  =  Pp  =  a  constant.       .    .     .     (i) 


Thus,  a  series  of  curves  may  be  obtained  in  which  p  varies 
inversely  as/,  and  the  hydrostatic  arch  is  that  curve  for  which 
the  pressure  p  at  any  point  is  directly  proportional  to  the  depth 
of  the  point  below  a  given  horizontal  plane. 

Denote  the  depth  by_j/.  and  let  iv  be  the  specific  weight  of 
the  substance  to  which  the  pressure/  is  due.     Then 


and 


p  =  wy, (2) 

T  =  Pp  =  wyp  =  a  constant.      •     •     •     (3) 
The  curve  may  be  delineated  by  means  of  the  equation 


yp  =  const. 


(4) 


!<■ 


h 


-t  >} 


iir' 


i'! 


■■,■«■ 


:)' 


758 


THEORY  OF  STRUCTURES. 


It  may  be  shown,  precisely  as  in  Case  {c\  that  the  horizontal 
intensity  of  pressure  {J>^ 


=  the  vertical  intensity  (Py)  =p. 


•    (5) 


Take  as  the  origin  of  co-ordinates  the  point  O  vertically 
above  the  crown  of  the  arch,  in  the  given  horizontal  plane. 
Let  the  horizontal  line  through  O  be  the  axis  of  x. 
"      "    vertical         "  "         "    "     "       "    "  j. 

Any  portion  AM  oi  the  arch  is  kept  in  equilibrium  by  the 
0  equal  thrusts  {T)  at  A  and  A/, 

and  by  the  resultant  load  P  upon 
AM,  which  must  necessarily  act 
in  a  direction  bisecting  the  angle 
ANM. 
Fio.  490.  Complete   the    parallelogram 

AM,  and  take  SIV  =  NM  to  represent  T. 

The  diagonal  NL  will  therefore  represent  P. 
Let  Q  be  the  inclination  of  the  tangent  at  M  to  the  hori- 
zontal. 

The  vertical  load  upon  AM=  vertical  component  of  P 

=  LK  =  Tsin  6  =  pfJt  sin  S  =  wyp  sin  6  =  wy^p^  sin  6,  .     (6) 

J/,,  p„  being  the  values  of  j,  p,  respectively,  at  A. 

The  horizontal  load  upon  A  AT  =  horizontal  component  oi  P 

=  NK=SN-KS=  T-  Tcosd=2Tism^ 


2f) 


2tvyp  sin  —  =  zwj/^p, 

4^ 


=  2pP  (sin  -j 
Again,  the  vertical  load  upon  AM 


••  (^'"  I)  • 


(7) 


=   /   pi^-^  —  "^^  J   ^'^^  ~  "^'^'Po  sin  ^ ; (8) 

the  horizontal  load  upon  AM 

^  X!^'^^  "^  ^X!^'^^  ~  ^^^'  ~-^'*^  ~  ^^yoP,  (sin  -)  .  (9) 


EXAMPLES  OF  LINEAR  ARCHES. 


759 


Equation  (8)  also  shows  that  the  area  bounded  by  the  curve 
AM,  the  verticals  through  M  and  A,  and  the  horizontal 
through  0  is  equal  to  yj\  sin  ^,  and  is  tlierefore  proportional 
to  sin  ^.  At  the  points  defined  hy  B  —  90°  the  tangents  to 
the  arch  are  vertical,  and  the  portion  of  the  arch  between  these 
tangents  is  alone  available  for  supporting  a  load.  The  vertical 
and  horizontal  loads  upon  one-half  the  arch  are  each  equal  to 

Corollary. — The  relation  given  in  eq.  (i)  holds  true  in  any 
arch  for  elements  upon  which  the  pressure  is  wholly  normal. 

This  has  been  already  proved  for  the  parabola  and  catenary, 
in  cases  {a)  and  (p). 

At  the  point  A'  of  the  elliptic  arch, 


c'r^ 


OA' 


=  —  =  cr. 


Hence,  the  horizontal  thrust  at  A' 


=  p,p--f>-pcr-  cH. 

{e)  Geostatic  Arch. — IhQ  geostatic  is  a  parallel  projection  of 
the  hydrostatic  arch. 

The  vertical  forces  and  the  lengths  of  vertical  lines  are 
unchanged.  '  _-— ^A 

The  horizontal  forces  and  lengths  of  hori- 
zontal  lines   are   changed   in  a   given  ratio 

^  to  I.  /B__yB: 

Let  B'A  be  the  half-geostatic  curve  de-  Fig.  491. 

rived  from  the  half-hydrostatic  curve  BA. 
The  vertical  load  on  AB' 


-P'  =  P=  thrust  along  arch  at  B'.  ...  (i) 
The  horizontal  load  on  AB' 

—  H'  ~  cH  =  thrust  along  arch  at  A.  .  .  ,  (2) 
The  new  vertical  intensity 


=  A' 


C.OB-  c    ~  c ^3i 


m 


%  m  \\ 


OB' 


\V  ili  M 


76p 


THEORY  OF  STRUCTURES. 


The  new  horizontal  intensity 

_      '  _  J^'  _  £^  _ 
~^'  ~  0A~  OA  ~'^^^ 


cp. 


(4) 


Thus,  the  geostatic  arch  is  suited  to  support  a  load  so  dis- 
tributed as  to  produce  at  any  point  a  pair  of  conjugate  press- 
ures;  pressures,  in  fact,  similar  to  those  developed  according 
to  the  theory  of  earthwork. 

Let  R^ ,  R^  be  the  radii  of  curvature  of  the  geostatic  arcli 
at  the  points  A,  B\  respectively,  and  let  r, ,  r,  be  the  radii  of 
curvature  at  the  corresponding  points  A,  B  o{  the  hydrostatic 
arch. 

The  load  is  wholly  normal  at  A  and  B' .     Thus, 

H'  =  p;R,  =  ^R,  =  cH=cpr,.    .     .    . 


Also, 


.-.  R,  =  c'r, 

P'  =:  p^'R,  =  cpR,  ^P  =  pr, 


(6) 
(7) 


cR,  =  r. 


(8) 


{f)  General  Case. — Let  the  figure  represent  any  linear 
P  arch  suited  to  support  a  load  which  is  sym- 
metrically distributed  with  respect  to  the 
crown  A,  and  which  produces  at  every  point 
of  the  arch  a  pair  of  conjugate  pressures, 
the  one  horizontal  and  the  other  vertical. 

Take  as  the  axis  oi y  the  vertical  through 
the  crown,  and  as  the  axis  of  x  the  hori- 
FiG.  49a.  zontal  through  an  origin  6?  at  a  given  dis- 

tance from  A. 

Any  portion  AM  oi  the  arch  is  kept  in  equilibrium  by  the 
horizontal  thrust  H  at  A,  the  tangential  thrust  T  at  M,  and 
the  resultant  load  upon  AM,  which  must  necessarily  act  through 
the  point  of  intersection  A'^  of  the  lines  of  action  of  H  and  T. 
Since  the  load  at  A  is  wholly  vertical,  H  is  given  by 


//■„  =  /  P« , 


(0 


(6) 

(7) 


Irough 
hori- 
bn  dis- 

the 
r,  and 
Irough 
Id  T. 


(0 


EXAMPLES  OF  LINEAR  ARCHES. 


761 


/>„  and  p„  being,  respectively,  the  vertical  intensity  of  pressure 
and  the  radius  of  curvature  at  A. 

Let  MN=  T,  and  take  NS  =  H,. 

Complete  the  parallelogram  SM\  the  diagonal  NL  is  the 
resultant  load  upon  AM  in  direction  and  magnitude. 

The  vertical  {KL)  and  the  horizontal  {KN)  projections  of 
NL  are,  therefore,  respectively,  the  vertical  and  horizontal 
loads  upon  AM. 

Denote  the  vertical  load  by  V,  the  horizontal  by  H.     Then 


7"  sin  0  =  KL  —  V—  J  pydx, (2) 


and 


H^  KN=  SN-  SK=H,-  Fcot  B,    .    .    (3) 
6  being  the  angle  between  MJV  and  the  horizon. 

dV 


Py,  the  vertical  intensity  of  pressure,  = 
px,  the  horizontal   intensity  of  pressure 


dx' 


dH  d  ,^_        ^, 

=  -7-=  -  -r(^'cot^). 
dy  d)r  ' 


(4) 


(5) 


Example. — A  semicircular  arch  of  radius  r,  with  a  hori- 
zontal extrados  at  a  vertical  distance  R  from  the  centre. 
The  angle  between  the  radius  to  J/ and  the  vertical  =  6. 

.'.x^rsxwB,    y  =  R  —  r  cos  0.    .     .    .    (i) 

'     dx  =r  cos  6d0,    dy  —  r  s,m  Odd (2) 

p^  =  wy  =  w{R  —  r  cos  (f), (3) 

w  being  the  specific  weight  of  the  load.     Hence, 

V=  u'J^\j^  -  r  cos  ey  cos  fidd 

=  wr[R  sm  &  -  — — J.    ...    (4) 


IF 

'i' 1 

I 
I 


m 


it-.' 
t  ■■ 


P 


762  THEORY  OF  STRUCTURES. 

Equations  (3)  and  (4)  give  H  \  iox 

/,  =  wiji  —  r) ,       .... 


and  hence 


H,  =  wr{R  -r) 

Px ,  the  horizontal  intensity  of  pressure, 

r  6  —  sin  6  cos  6 


(5) 

(6) 


=  -~{Vcote)=^.[R-l' 


sin  0 


—  r  cos  B 


)•   (7) 


Rankine  gives  the  following  method  of  determining  whether 
a  linear  arch  may  be  adopted  as  the  intrados  of  a  real  arch. 
At  the  crown  a  of  a  linear  arch  ab  measure  on  the  normal  a 
length  ac,  so  that  c  may  fall  within  the  limits  required  for 
stability  (e.g.,  within  th^  middle  third). 

At  c  two  equal  and  opposite  forces,  of  the  same  magnitude 
as  the  horizontal  thrust  H  at  a,  and  acting  at  right  angles  to 
ac,  may  be  introduced  without  altering  the  equilibrium. 

Thus  the  thrust  at  a  is  replaced  by  an  equal  thrust  at  c,  and 
a  right-handed  couple  oi  moment  H .  ac. 

Similarly,  the  tangential  thrust  T'at  any  point  d  of  ab 
ma}'  be  replaced  by  an  equal  and  parallel  thrust  at  e,  and  a 
couple  of  moment  T .  de. 

The  arch  will  be  stable  if  the  length  of  de,  which  is  normal 
to  ab  at  d,  is  fixed  by  the  condition  7".  de  ^^  H .  ac,  and  if  the 
line  which  is  the  locus  of  e  falls  within  a  certain  area  (e.g., 
within  the  middle  third  of  the  arch  ring. 

8.  Arched  Ribs  in  Iron,  Steel,  or  Timber. — In  the  fol- 
lowing articles,  the  term  arched  rib  is  applied  to  arches  con- 
structed of  iron,  steel,  or  timber.  The  coefficients  of  elasticity 
are  known  quantities  which  are  severally  found  to  lie  between 
certain  not  very  wide  limits,  and  their  values  maybe  introduced 
into  the  calculations  with  the  result  of  giving  to  them  greater 
accuracy.  There  are  other  considerations,  however,  involved 
in  the  problem  of  the  stability  of  arched  ribs  which  still  render 
its  solution  more  or  less  indeterminate. 

It  has  been  shown  that  the  curve  of  pressure,  or  linear  arch, 


m-e^.  \ 


ARCHED   RIB    U.VDER  A     VERTICAL   LOAD. 


763 


is  a  funicular  polygon  of  the  extraneous  forces  which  act  upon 
the  real  arch.  It  is,  therefore,  also  the  bending-moinent  curve, 
drawn  to  a  definite  scale,  for  a  similarly  loaded  Jiorizontal 
girder  of  the  same  span,  whose  axis  is  the  springing  line. 

When  the  arched  rib  carries  a  given  symmetrically  dis- 
tributed load,  it  will  be  assumed  that  the  linear  arch  coincides 
with  the  axis  of  the  rib,  and  that  the  thrust  at  any  normal 
cross-section  is  axial  and  uniformly  distributed. 

The  total  stress  at  any  point  is  made  up  of  a  number  of 
subsidiary  stresses,  of  which  the  most  important  are  :  (l)  a 
direct  thrust ;  (2)  a  stress  due  to  flexure ;  (3)  a  stress  due  to  a 
change  of  temperature.  Each  of  these  may  be  investigated 
separately,  and  the  results  superposed. 

9.  Bending  Moment  (M)  and  Thrust  (T)  at  any  Point 
of  an  Arched  Rib  under  a  Vertical  Load. — Let  ABC  be  the 
axis  of  the  rib. 

Let  D  and  E  be  points  on  the  same  vertical  line,  E  being 

'D^ 


Fig.  493. 

on  the  axis  of  the  rib  and  D  on  the  linear  arch  for  nny  given 
distribution  of  load. 

Resolve  the  reaction  at  A  into  its  vertical  and  horizontal 
components,  and  denote  the  latter  by  H, 

Since  all  the  forces,  excepting  H,  are  vertical,  the  difference 
between  the  moments  at  D  and  E  =  H .  DE. 

But  moment  at  Z>  =  o.     Hence, 

moment  at  i?  =  J/=  H .  DE. 

Let  the  normal  at  E  meet  the  linear  arch  in  D'.  Then,  if 
T  is  the  thrust  along  the  axis  at  E, 

J)'E 
T  cos  DED'  —  H  =  ^^p^  approximately, 


or 


;il 


1!  ■ 


H.DE--^  T.D'E=  M. 


764 


THEORY  OF  STRUCTURES. 


10.   Rib  with   Hinged  Ends ;  Invariability  of  Span.— 

Let  ABC  be  the  axis  of  a  rib  supported  at  the  ends  on  pins  or 


--.^.-^.-^8 


Fig.  494. 


i 


M 


on  cylindrical  bearings.  The  resultant  thrusts  at  A  and  C 
must  necessarily  pass  through  the  centres  of  rotation.  The 
vertical  components  of  the  thrusts  are  equal  to  the  corre- 
sponding reactions  at  the  ends  of  a  girder  of  the  same  span 
and  similarly  loaded,  and  H  is  given  by  the  last  equation  in 
the  preceding  article  when  BE  has  been  found. 

Let  ADC  be  the  linear  arch  for  any  arbitrary  distribution 
of  the  load,  and  let  it  intersect  the  axis  of  the  rib  at  6^  Tlie 
curvature  of  the  more  heavily  loaded  portion  AES  will  be 
flattened,  while  that  of  the  remainder  will  be  sharpened. 

The  bending  moment  at  any  point  E  of  the  axis  tends  to 
change  the  inclination  of  the  rib  at  that  point. 

Let  the  vertical  through  E  intersect  the  linear  arch  in  D 
and  the  horizontal  through  A  in  F. 

Let  6  be  the  inclination  of  the  tangent  at  £  to  the  hori- 
zontal. 

Let  /  be  the  moment  of  inertia  of  the  section  of  the  rib 
at£. 

Let  ds  be  an  element  of  the  axis  at  E. 

Mds      H.DE.ds 


Change  of  inclination  3i^  E  =  dO  —  —=j 


EI 


If  this  change  of  curvature  were  effected  by  causing  the 
whole  curve  on  the  left  of  E  to  turn  about  E  through  an  angle 
dd,  the  horizontal  displacement  of  A  would  be 


EF.  dO  = 


?*  . 


H.DE.EF.ds 
EI 


ARCHED  RIB    WITH  HINGED  ENDS. 


765 


This  is  evidently  equal  to  the  horizontal  displacement  of 
E,  and  the  algebraic  sum  of  the  horizontal  displacements  of  all 
points  along  the  axis  is 


M.DE.EF.ds 


=/^ 


DE.EF.ds 


EI 


=  0, 


(I) 


since  the  length  AC  '\s  assumed  to  be  invariable. 

Thus,  the  actual  linear  arch  must  fulfil  the  condition  ex- 
pressed by  eq.  (i),  which  may  be  written 


J 


'DE.EF.ds 


=  0, 


(2) 


since  H  and  E  are  constant. 

If  the  rib  is  of  uniform  section,  /is  also  constant,  and  eq.  (2) 
becomes 

jDE.EF.ds=o (3) 

Also,  since  DE  is  the  difference  between  DF  and  EF, 

C{pF  -  EF)EF.  ds  =  o  =fDF.  EF .  ds- J EF'ds  (4) 

Remark. — Eq.  i  expresses  the  fact  that  the  span  remains 
invariable  when  a  series  of  bending  moments,  H .DE,  act  at 
points  along  the  rib.  These,  however,  are  accompanied  by  a 
thrust  along  the  arch,  and  the  axis  of  the  rib  varies  in  length 
with  the  variation  of  thrust. 

Let  H^  be  the  horizontal  thrust  for  that  symmetrical  loading 
which  makes  the  linear  arch  coincide  with  the  axis  of  the  rib. 

Let  T^  be  the  corresponding  thrust  along  the  rib  at  E. 

The  shortening  of  the  element  ds  at  E  of  unit  section 


T 


-ds. 


Example  i.  Let  the  axis  of  a  rib  of  uniform  section  and 
hinged  at  both  ends  be  a  semicircle  of  radius  r. 

Let  a  single  weight  W  be  placed  at  a  point  upon  the  rib 
whose  horizontal  distance  from  O,  the  centre  of  the  span,  is  a. 


%\-  jg  11     I 


111 


1.^'    IH 


( 


iAs.^a 


itii^-i 


^(A 


THEORY  OF  STRUCTURES. 


The  "  linear  arch  "  (or  bending-moment  curve)  consists  of 
two  straight  lines  DA,  DC. 


/  \\ 


Draw  any  vertical  line  intersecting  the  axis,  the  linear 
arch,  and  the  springing  line  AC  in  E\  D',  F',  respectively. 

Let  OF'  =  X,  and  let  dx  be  the  horizontal  projection  upon 
AC  ol  the  element  ds  at  E'. 

Then 

ds  T^,^r-,  ^ 

^  =  cosec  E'OF  =  -^j, , 


or 


E'F'ds  =  rdx (i) 


Applying  condition  (4), 

f^  D'F'rdx  -f  f  D'F'rdx  =  f  E'F'rdx, 


or 


y"  D'F'dx  +  f  D'F'dx  =  f  E'F'dx, 

or  area  of  triangle  ADC  —  area  of  semicircle. 

And  if  z  be  the  vertical  distance  of  D  from  AC, 


nr' 


zr  = 


;f  r 


or 


ARCHED   RIB    WITH  HINGED   ENDS,  j6y 


nr 
z  =■  —  —  one-half  of  length  of  rib.     ...     (2) 


I       $: 


nr 


.:  DE  =  DF-  EF=   ~-  Vr*  -  a' 

2 


(3) 


Hence,  if  h  be  the  horizontal  thrust  on  the  arch  due  to  W, 

3  3 

h .  DE  =  M  =  W^-^^^ (4) 

Similarly,  if  there  are  a  number  of  weights  W^,  IV,,  W^, . .  . 
upon  the  rib,  and  if  //,,  ^,,  ^3,  .  .  .  are  the  corresponding  hori- 
zontal thrusts,  the  total  horizontal  thrust  H  will  be  the  sum  of 
these  separate  thrusts,  i.e., 


/f=//.  +  /^,+ 


(5) 


It  will  be  observed  that  the  apices  (Z>, ,  D^,  D^,  .  .  .)  of  the 

several  linear  arches  (triangles)  lie  in  a  horizontal  line  at  the 

nr 
vertical  distance  —  from  the  springing  line. 

Ex.  2.  An  ar'-.hcd  rib  hinged  at  the  ends  and  loaded  with 
weights  W^,  W,,  W,,  .  .  . 


\ 
\ 
\ 

0  ^i^^v,  s 


r: 


Fig.  4g6. 


l^ 
Fig.  497. 


Let  I  2  3  4  .  .  .  «  be  the  line  of  loads,  W^  being  represented 
by  I  2,  W^  by  2  3,  W^   by  34,  etc..  and  let  the  segments   \Xy 


i!|i;    k:  \'      \ 


LIU  \ 


768 


THEORY  GF  S'JRUCTURES. 


nx,  respectively,  represent  the  vertical  reactions  at  A  and  C. 
Take  the  horizontal  length  xPto  represent  //,  and  draw  the 
radial  lines  Pi,  P2,  Pi,  .  .  . 

The  equilibrium  polygon  ^^'-.i'-^,  .  .  .  must  be  the  funicu- 
lar  polygon  of  the  forces  with  respect  to  the  pole  P,  and  there- 
fore the  directions  of  the  resultant  thrusts  from  A  to  Z:',,  £^  to 
/;,,,  P^  to  /ij,  ...  are  respectively  parallel  to  Pi,  P2,  P^,  .  .  . 

The  tangential  (axial)  thrust  and  shear  at  any  point  />  uf 
the  rib,  e.g.,  between  Zi,  and  £,,  may  be  easily  found  by  draw- 
ing Pi  parallel  to  the  tangent  at/,  and  3/  perpendicular  to  J'/. 
The  direct  tangential  thrust  is  evidently  represented  by  P/, 
and  the  normal  shear  at  the  same  point  by  3/.  The  latter  is 
borne  by  the  web. 

If  /  is  a  point  at  which  a  weight  is  concentrated,  e.g.,  £^, 
draw  /V7"  parallel  to  the  tangent  at  £,  and  5/',  6i"  perpen- 
dicular to  Pi'i". 

Pt'  represents  the  axial  thrust  immediately  on  the  left  of 
^5 ,  and  5/' the  corresponding  normal  shear,  while /^^"  repre- 
sents the  axial  thrust  immediately  on  the  right  of  E^,  and  6/" 
the  corresponding  normal  shear. 

A  vertical  line  through  P  can  only  meet  the  line  of  loads 
at  infinity. 

Thus,  it  would  require  the  loads  at  A  and  C  to  be  infinitely 
great  in  order  that  the  thrusts  at  these  points  might  be  vertical. 
Practically,  no  linear  arch  will  even  approximately  coincide 
with  the  axis  of  a  rib  rising  vertically  at  the  springings,  and 
hence  neither  a  semicircular  nor  a  semi-elliptical  axis  is  to  be 
recommended. 

Ex.  3.  Let  the  axis  of  the  rib  be  a  circular  arc  of  span  2/ 
and  radius  r,  subtending  an  angle  201  at  the  centre  N. 

Let  the  angles  between  the  radii  NE,  NE'  and  the  vertical 
be  p  and  0,  respectively. 

The  element  ds  at  E'  =  rdO. 

Also,  E'F'  =  r(cos  d  -  co^  a);    AF  =  /-r  sin  d- 


D'F'  ^  r,—n  -  r  sin  B). 


ARCHED   RIB    WITH  HINGED   ENDS, 


769 


yo- 


N 

Fic.  498. 


Applying  condition  (5), 

r'(cos  ^  —  cos  afrdO 


L 


=/ 


7— — (/— r  sin  ^)r(cos  0  —  cos  a)rd9 


+ 


X^-^/- 


r  sin  6)r{cos  S  —  cos  a)rdO, 


which  easily  reduces  to 

r{ar(cos  2a  +  2)  —  f  sin  2a\ 
=  7, i  \  /'(sin  oc  —  a  cos  «)  +  -  (cos  2a  —  cos  2^) 

—  /■/  cos  a'(cos  a  —  cos  /5)  —  /«(sin  /?  —  /?  cos  a)  >• , 

an  equation  giving  z  or  Z?/^     Also, 

DE  =  DF-  EF, 

and  the  corresponding   horizontal  thrust  may  be  found,  as 
before,  by  the  equation 


h,DE=  W 


I    f 


il.  M 


L 


770  THEOKY  OF  STRUCTURES. 

Note.—U  a°  =  90°, 


5  _    2"    (^  ~  ^*] 


or     £  ~  —  as  in  Ex.  I. 


Ex.  4.     Let  the  axis  be  a  parabola  of  span  2/  and  rise  Jt. 
(Fig.  498,  Ex.  3),     From  the  properties  of  the  parabola, 


E'F' 


=  /&(i-^). 


t  ±  a 


and 


or,  approximately. 


ds*  =  «'.ir'(i  +  4^^'), 


+4V). 


^j  =  dx\  I 


Applying  condition  (5), 

which  easily  reduces  to 

an  equation  giving  2  or  Z>K 

iVc^/^. — If  the  arch  is  very  flat,  so  that  ds  may  be  considered 


ARCHED  RIH    WITH  ENDS  ABSOLUTELY  FIXED.       77 ^ 


/{•' 


as  approximately  equal  to    ix,  the  term  2j,x'  in   the  above 
equation  may  be  disregarded,  and  it  may  be  easily  shown  that 


I  /'  +  a' 


{■-a-q 


i6 

'is' 


or 


kr 


•      -  -  5  5/'  -  a" 

II.  Rib  with  Ends  absolutely  Fixed.— Let  ABC  be  the 
axis  of  the  rib.     The  fixture  of  the  ends  introduces  two  un- 


Fig.  499. 

known  moments  at  these  points,  and  since  H  is  also  unknown, 
three  conditions  must  be  satisfied  before  the  strength  of  the 
rib  can  be  calculated. 

Represent  the  linear  arch  by  the  dotted  lines  KL ;  the 
points  K,  L  may  fall  above  or  below  the  points  A,  C. 

Let  a  vertical  line  DEF  intersect  the  linear  arch  in  D,  the 

axis  of  the  rib  in  E,  and  the  horizontal  through  A  in  F. 

r   .     ,.       .  ^  ,„       Mds 

As  m  Art.  lo,  change  of  mclination  at  h,  or  dB,  =  —f^. 

k.1 

But  the  total  change  of  inclination  of  the  rib  between  A  and 

C  must  be  nil,  as  the  ends  are  fixed. 


pMds 
'J  ~EI 


°  =  / 


H.DE.ds 
EI       ' 


(I) 


■which  may  be  written 


/ 


-j-ds  =  O, 


(2) 


II 


tf 


since  H  and  E  are  constant. 


ibiv     ! 


U 


772 


THEORY  OF  STRUCTURE'S. 


If  the  section  of  the  rib  is  uniform,  /  is  constant  and  eq. 
(2)  becomes       .  v*'- •"  i'---  .-•  •     v   •  •• 


/ 


DE  .ds  =  0. 


(3) 


Again,  the  total  horizontal  displacement  between  A  and  C 
will  be  nil  if  the  abutments  are  immovable.  If  they  yield,  the 
amount  of  the  yielding  must  be  determined  in  each  case,  and 
may  be  denoted  by  an  expression  of  the  form  f^H,  M  being 
some  coefificient.  ..  ._"'■.    ■    .: 

As  in  Art.  10,  the  total  horizontal  displacement        l    .'   .; 


H.DE.EF.ds 


•••/ 


~  J  EI 

H.DE.EF.ds 


EI 


=  o    or     =  jxH.   ...    (4) 


But  H  and  E  are  constant. 

>DE .  EF.  ds 


•■•/- 


/ 


=  o    or     = 


E' 


.    (5) 


If  the  section  of  the  rib  is  uniform,  /  is  also  constant,  and 
hence 


/ 


DE  .  EF  .ds  =  o    or 


E' 


...    (6) 


and  since  DE  is  the  difference  between  DF  and  EF,  this  last 
may  be  written 

^fDF.EF.ds  '^J'EF\ds  =  o    or    =^I.     ...     (7) 

Again,  the  total  vertical  displacement  between  A  and  C 
must  be  nil. 
._.  The  vertical  displacement  of  E  (see  Art.  10) 

M.AF.ds 


-AF,dB^ 


Ml 


ARCHED   RIB    WITH  ENDS  ABSOLUTELY  FIXED.       773 

Hence,  the  total  vertical  displacement 

^H.DE.AF 


f- 


EI 


ds  =  0, (8) 


which  may  be  written 


-DE .  AF 


/ut. .  At  , 
J — as  —  o, 


(9) 


since  //'and  E  are  constant.     If  the  section  of  the  rib  is  also 
constant, 

fDE.AF.ds=o=fDF,AF.ds-fEF,AF.ds.    (lo) 

t 

Eqs.  (2),  (5),  and  (9)  are  the  three  equations  of  condition. 

In  eq.  (9)  AF  must  be  measured  from  same  abutment 
throughout  the  summation. 

The  integration  extends  from  A  to  C. 

Example  i.  Let  the  axis  of  the  rib  be  a  circular  arc  of 
span  2/,  subtending  an  angle  2a  at  the  centre  N'. 

Let  a  weight  IV ,  be  concentrated  on  the  rib  at  a  point  E 


!       y 


^.2  ft  / 

N 

Fig,  500. 

whose  horizontal  distance  from  the  middle  point  of  the  span 
is  a. 

Let  the  radius  NE  make  an  angle  /?  with  the  vertical. 
.    The  "  linear  arch  "  consists  of  two  str-ight  lines  DA',  DC. 

Let  AA'  —  y,.  DF  =  z,  CC  =)\ . 


I' 


I    J 


5       f< 


774  \  THEORY  OF  STRUCTURES. 

Draw  any  ordinate  E'F'  intersecting  the  linear  arch  in  Lf. 
Let  the  radius  NE'  make  an  angle  ^  with  the  vertical. 
Then 

E'F'  =  r{cos  6  —  cos  a). 


AF'  =  /-rsme,    and    D'F' =={/ -r sine)'^—^-{.y^ 


if  F'  is  on  the  left  of  F; 


AF'  =  I -\-r  sine    and     D'F' =  {/- r  sin  d)-j-^  J^  y^ 


l-a 


\{  F'  is  on  the  right  of  F. 

Also,  ds  =  rdB.  i 

Applying  condition  (i), 

+  r  {('-'•  sin  «)^|^ +>.,  I  </# 

=  r  /  (cos  d  —  cos  o()dB (i) 

Applying  condition  (3),  and  assuming  M  =  0, 

I    {cos  e-cosa)U/-r  sin  ^Oj^^'  +^.  N^ 

H-   Acos  ^  -  cos  «)  I  (/- r  sin  ^^^*  +  J,  I 
=  r/   (cos  6^  —  cos  a')V^ (2) 


ARCHED  RIB    WITH  ENDS  ABSOLUTELY  FIXED.       775 
Applying  condition  (5), 

•  A/ -  r sin  e)Ul-r  sin  ^fj~^  +  J.  [  ^^ 

H-   A/+  r  sin  B)  i  (/-  r  sin  ef^^J^y^  \  dO 

=  r  I  (cos  (f  —  cos  «)(/  —  rsin  6)dd 

* 

4-  rj"" {cos  e  —  cos  «)(/  +  r  sin  ^^^.     (3) 

Equations  (i),  (2),  (3)  may  be  easily  integrated,  and  the  re- 
sulting equations  will  give  the  values  of  7, ,  s,  andj, . 

The  corresponding  horizontal  thrust,  h,  may  now  be  ob- 
tained from  the  equation  h  .  DE  =.  M  =  h{z  —  EF). 

Note. — If  the  axis  is  a  semicircle,  and  if  W^is  at  the  crown, 

a  =^  O,     a  =  90",     yS  =  o, 
and  eqs.  (i),  (2),  (3)  reduce  to 

5(7r  -  2)  +  j/,(^  -  i)  -  j/,(^  +  i)  =  2r. 

n  —  2  ,  4+  2;r  —  ?r* 

.\z=r ,    and    y^  =  y^=zr . 

4  —  TT  •!       •»  4  —  w 

Ex.  2.   Let  the  axis  be  a  parabola  of  span  2/  and  rise  k 
(Fig.  500  in  Ex.  i). 
.A.S  in  Ex.  3,  Art.  10, 

E'F'  =  i{i  -  pj;     ds'  =  (i  -f  ^V)^;ir. 


i  I'l'^ 


776 
Also, 

and 


THEORY  OF  STRUCTURES. 


D'F'  =7,  +  {l~xYj-^}  on  the  right  of  DF, 


'  l-\-a 


:  J,  +  (/  —  X)- — --  on  the  left  of  DP, 


AF'  =  IT  X. 
The  equations  of  condition  become 


Thej 
ing  equj 

If  th 
proxima 


12.  E 

in  the  sj 
temperat 
expansio 
Henci 
temperat 
expressec 


If  the 
E  is  also  ( 


EXAMI 


be  the  arc  ( 
centre. 
First,  le 


EFFECT   OF  A    CHANGE    OF    TEMPERA  T UK E 


777 


These  equations  may  be  at  once  integrated,  and  the  result- 
ing equations  will  give  the  values  of  j, ,  j,j,  s. 

If  the  arch  is  very  flat,  so  that  ds  may  be  taken  to  be  ap- 
proximately the  same  as  lix,  it  may  be  easily  shown  that; 

2/4-5^  2    I  —  t^a  ,  6, 

^i  =  ~:""TT" — »    yt  —  ~7~/ »     and    z  =  -k. 

-^*       li,    l-\-a     -'^       15/— rt'.  5 

12.  Effect  of  a  Change  of  Temperature. — The  variation 
in  the  span  2/  of  an  arch  for  a  change  of  t°  from  the  mean 
temperature  is  approximately  =  2etl,  e  being  the  coefificient  of 
expansion. 

Hence,  if  Ht  is  the  liorizontal  force  induced  by  a  change  of 
temperature,  the  condition  that  the  length  AC  is  invariable  is 
expressed  by  the  equation 


"■f 


DE.EF.ds 
EI 


±  26//  =0. 


If  the  rib  is  of  uniform  section,  /  is  constant ;  and  since 
E  is  also  constant,  the  equation  may  be  written 


~-  j  DE.  EF.  (is  ±  2etl  =  o. 


Example  i.  Let  the  axis  AEC  o(  a  rib  of  uniform  section 


N 


Fig.  501. 


be  the  arc  of  a  circle  of  radius  r  subtending  an  angle  2a  at  the 
centre. 

Firsi,  let  the  rib  be  hinged  at  bolh  ends. 


: :  W: 


mk 


7;8 


THEORY  OF  STRUCTURES. 


It  is  evident  that  the  straight  line  AC\?>  the  "linear  arch." 
Then,  - 

fDE .  EF.  ds  =zj*EF'ds  =  r'  J\cos  6  -  cos  afdd 

=  r'|a(2-f  cos  2a)  —  |sin2a'|. 
Also,  /  =  r  sin  a. 

''•^iin^^'*^^+^°^^"^~*^'"^"^  ±26//=  a 

Note, — If  the  axis  is  a  semicircle,  a  =  90°,  and 

.—  ~±2etl:=:0. 

Second,  let  the  rib  h^  fixed  at  both  ends. 
The  "  linear  arch"  is  now  a  straight  line  A'C  at  a  distance 
j?(=    DF)  !"   ;-n  .<4C  given  by  the  equation 

/  DE .ds  =  o. 
.'.  JdF  .  ds  =fEF .  ds, 

e  ids  =  r'  I   (cos  d  —  cos  a)dB, 
az  =  r(sin  a  —  a  cos  a). 


or 

or 

Also, 


J' BE .  EF.  ds  ^f{DF.  EF-^EPys  =  zjEFds  -^fEF'ds 
—  2^rr'(sin  or  —  «  cos  «)~  r'|«(2  +  cos  2a)  —  |  sin  2a\. 

.|  2^r'(sin  a  —  or  cos  a)  —  r'|a(2  +  cos  2a)  —  |  sin  2a  |  ( 

±  2etl  =  o, 


EI 


and    /  =  r  sin  a. 


Ex.  2.  Let  the  axis  AEC  of  a  rib  of  uniform  section  be  a 
parabola  of  span  2/ and  rise  k.     (See  Fig.  501  in  Ex.  i.) 


/ 


Fin 
The 

DE 


and  hen 


Secon 
The 
from  AC 


or 


.'.  ^ 


or 


Also, 
fDE.  EI 

=zfEF.i 


779 


EFFECT  OF  A    CHANGE   OF   TEMPERATURE. 

First,  let  the  rib  be  hinged  at  both  ends. 

The  straight  line  AC\s  the  linear  arch.     Then 


and  hence, 


~  k%  -;  4-  --—  77     ±  2€tl  =  O. 

EI     V15  '  105 /V 


or 


Second,  let  the  rib  ht  fixed  eA.  both  ends. 
The  linear  arch  is  the  line  A'C  at  a  distance  z  {z=.DF) 
from  A  C  given  by  the  equation 


CdE ,ds  =  o  =  J{DF ~  EF)ds, 
DFCds^  fEF.ds. 


or 


Also, 


/"/?£  .EF.ds  =fDF  .EF.ds-^  J  EF'ds 


i 


ii- 


ft     1 


I  li 


78o 


THEORY  OF   STRUCTURES. 


Hence, 


Ht  Akl  {     I     .2k 


{.(.+-;f;)~f(.+f^)} 


±  2etl  =  o. 


Remark. — The  coefficient  of  expansion  per  degree  of  Fah. 
renheit  is  .0000062  and  .0000067  for  cast-  and  wrought-iron 
beams,  respeictively.  Hence,  the  corresponding  total  expansion 
or  contraction  in  a  length  of  100  ft.,  for  a  range  of  60°  F.  from 
the  mean  temperature,  is  .0372  ft.  {—  -^a")  and  .0402  ft.  (=  ^"). 

In  practice  the  actual  variation  of  length  rarely  exceeds  (?«r- 
half  oi  these  amounts,  which  is  chiefly  owing  to  structural  con- 
straint. 

13.  Deflection  of  an  Arched  Rib. 


Fig.  502. 

Let  the  abutments  be  immovable. 

Let  ABC  be  the  axis  of  the  rib  in  its  normal  position. 

Let  ADC  represent  the  position  of  the  axis  when  the  rib  is 
loaded. 

Let  BDF  be  the  ordinate  at  the  centre  of  the  span  ;  join 
AB,  AD. 

Then 

/arc  A  D\ ' 

DF'  =  AD'  -  AF'  =  AB'l ~\  -  AF". 

\arc  ABI 


But 


arc  AB  —  arc  AD      f 
arc  AB  ~E' 


/being  the  intensity  of  stress  due  to  the  change  in  the  length 
of  the  axis. 


/ 


.-.  DF'  =  AB'[i  -  ^-1  -  AF'  =  BF'  -  AB'  |  2^ 


E       \EI   ^ 


i 


ELEMENTARY  DEFORMATION  OF  AN  ARCHED  RIB.     78 1 

/.  AB'^  \^~e~^h)  1  "^  ^'^'~  ^^'  ^  ^^^^  ~  DF){BF+  DF) 

=  2BP\BD),  approximately. 
y—rj    is  also  sufficiently  small  to  be  disregarded.     Hence, 

AB'  f       k'  -{-r  f  .  , 

BD,  the  deflection,  =  -j^  jr  —  — 7 —  ^  ,  approximately. 

14.  Elementary  Deformation  of  an  Arched  Rib. 

X  0 


The  arched  rib  represented  by  Fig.  503  springs  from  two 
abutments  and  is  under  a  vertical  load.  The  neutral  axis  PQ 
is  the  locus  of  the  centres  of  gravity  of  all  the  cross-sections  of 
the  rib,  and  may  be  regarded  as  a  linear  arch,  to  which  the 
conditions  governing  the  equilibrium  of  the  rib  are  equally  ap- 
plicable. 

Let  AA'  be  any  cross-section  of  the  rib.  The  segment 
AA'P  is  kept  in  equilibrium  by  the  external  forces  which  act 
upon  it,  and  by  the  molecular  action  at  AA'. 

The  external  forces  are  reducible  to  a  single  force  at  C  and 
to  a  couple  of  which  the  moment  M  is  the  algebraic  sum  of 
the  moments  with  respect  to  C  of  all  the  forces  on  the  right 
of  C. 

The  single  force  at  C  may  be  resolved  into  a  component  T 
along  the  neutral  axis,  and  a  component  6"  in  the  plane  AA'. 


r<ii 


est 


y  ii 


It.  :i 


782 


THEORY  OF  STRUCTURES. 


The  latter  has  very  little  effect  upon  the  curvature  of  the  neu- 
tral axis,  and  may  be  disregarded  as  compared  with  M. 

Before  deformation  let  the  consecutive  cross-sectfons  BB' 
and  A  A'  meet  m  R\  R  is  the  centre  of  curvature  of  the  arc 
CC  of  the  neutral  axis. 

After  deformation  it  may  be  assumed  that  the  plane  AA* 
remains  unchanged,  but  that  the  plane  BB'  takes  the  position 
B"B"'.  Let  AA'  and  B"B"'  meet  in  R' ;  R'  is  the  centre  of 
curvature  of  the  arc  CC  after  deformation. 

Let  abc  be  any  layer  at  a  distance  z  from  C. 

Let  CC  =  As,  CR  =  R,  CR'  =  R\  and  let  Aa  be  the  sec- 
tional area  of  the  layer  adc. 

By  similar  figures, 

ac       R'-\-2  ^     ah      R-\-z 


.*.  be  ■=  ac  —  ab  ■=■  As .  z 


\R'  ~r)' 


The  tensile  stress  in  abc 


be  As., 

=  £  .  Aa—j  =  E  .  Aa j- 

ab  ab 


\R'       RJ 


2\^  —  -^],  very  nearly. 


=  E.Aa 
The  moment  of  this  stress  with  respect  to  C 

=  E .  Aa. 
Hence,  the  moment  of  resistance  at  AA' 


I 

R^ 


I 
R 


=Je  .  Aa  .  zi^  -  1)  =  e{^  -  -^)jAa .  A 
the  integral  extending  over  the  whole  of  the  section. 


ELEMENTARY  DEFORMATION  OF  AN  ARCHED  RIB.     T^l 


Again,  the  effect  of  the  force   T  is  to  lengthen  or  shorten 
the  element  CC\  so  that  the  plane  BB'  will  receive  a  motion 
of  translation,  but  the  position  of  R  is  practically  unaltered. 
Corollary  i.  Let  A  be  the  area  of  the  section  AA' , 
"   The  total  unit  stress  in  the  layer  abc 


T      Mz 


(2) 


the  sign  being  plus  or  minus  according  as  iT/acts  towards  or 
from  the  edge  of  the  rib  under  consideration. 

From  this  expression  may  be  deduced  (i)  the  position  of 
the  point  at  which  the  intensity  of  the  stress  is  a  maximum  for 
any  given  distribution  of  the  load ;  (2)  the  distribution  of  the 
load  that  makes  the  intensity  an  absolute  maximum  ;  (3)  the 
value  of  the  intensity. 

Cor.  2.  Let  w  be  the  total  intensity  of  the  vertical  load  per 
horizontal  unit  of  length. 

Let  7y,  be  the  portion  of  w  which  produces  only  a  direct 
compression. 

Let  H  be  the  horizontal  thrust  of  the  arch. 

Let  P  be  the  total  load  between  the  crown  and  A  A'  which 
produces  compression. 

Refer  the  rib  to  the  horizontal  OX  and  the  vertical  OPY 
as  the  axes  of  x  and  y,  respectively. 

Let  X,  y  be  the  co-ordinates  of  C, 

Then 


P="% 


but    dP  —  w^dx. 


also, 


dx 


1 

)- 

(5) 

(4) 

; 
[: 

} 

i 
I 

784 


rilEORV  OF  STNUCTUA'ES. 


15.  General  Equations. 

Let  /  be  the  span  of  the  arch. 

Let  X,  y  be  the  co-ordinates  of  the  point  C  ^</(?r^  deforma- 
tion. 

Let  x' ,  y'  be  the  co-ordinates  of  the  point  C  after  deforma- 
tion. 

Let  B  be  the  angle  between  tangent  at  Cand  6>^  te 

deformation. 

Let  B'  be  the  angle  between  tangent  at  C  and  OX  after 
deformation. 

Let  ds  be  the  length  of  the  element  CC  before  deforma- 
tion. 

Let  ds  be  the  length  of  the  element  CC  after  deformation. 

^i^'        I  ^     dd         \ 

Effect  of  flcxicre.     ^  —  ^       and     ^  =  j^- 

M      I       I     de'     de     dO'-dB 

'''Ei=R'~R  =  di'--di  =  —dr^'''''y''^^'^y' 

Let  i  be  the  change  of  slope  at  C.    Then 

Mds       Mds 
di  =  de-  dO'  =  Vf  =  4r,:7-^.r. 
LI        Eldx 


.:  t'  =  6-6'  =  i  -)- 


/ 


E/dx 


dx,      ...    (5) 


i^  being  the  change  of  slope  at  P,  and  a  quantity  whose  value 
has  yet  to  be  determined. 

Again,  the  general  equations  of  equilibrium  at  the  plane 
AA'  are 

d'M     dS  .  .  (         „dy\  ..^ 

for  the  portion  7«/, ,  Cor.  2,  Art.  14,  produces  compression  only 
and  no  shear. 

...5=s._  r«-rf.+^(l-|-j (7) 


GENERAL   EQUATIONS. 


785 


5,  beiny  the  still  undetermined  vertical  component  of  the  shear 

(iv 
at  P,  and   ■[-   the  slope  at  /'.     Also, 

M  =  M,  +  S,x  -  fjfj'^vdx^  ■VH[y-y,-  /j^'),    (8) 

Af^  being  the  still  undetermined  bending  moment  at  P. 

Equations  (5),  (6),  ("),  and  (8)  contain  the  /our  undeter- 
mined constants  //,  5„ ,  M, ,  z„ . 

Let  A/, ,  5, ,  and  i\  be  the  values  of  M,  S,  and  t,  respectively, 

at  (2- 

Equations  of  Condition. — In  practice  the  ends  of  the  rib  are 
either  ^.i7v/  or  free. 

If  they  are  fixed,  /„  =  o ;  if  they  are  free,  Af^  —  o.  In  either 
case  the  number  of  undetermined  constants  reduces  to  t/iree. 

If  the  abutments  arc  immovable,  ;r,  -  /  =  o.  If  the  abut- 
ments yield,  ,v,  —  /must  be  found  by  experiment.  Let  x^  —  I 
=  i-iH,  fJi  being  some  coefficient.  Hh^  first  equation  of  condi- 
tion is 

x^  — 1  =  0,     or     r,  —  /=  uH.    ....     (9) 

Again,  Q  is  immovable  in  a  vertical  direction,  and  the 
second  equation  of  condition  is 


y.-yo  =  o. 


(10) 


Again,  if  the  end  Q  is  fixed,  z,  =  O ;  and  if  free,  Af^  =0]  and 
the  t/drd  equation  of  condition  is 

i,=o,orAf,=o (11) 

Substituting'  in  equations  (7)  and  (8)  the  values  of  the  three 
constants  as  determined  by  these  conditions,  the  shearing  force 
and  bending  moment  may  be  found  at  any  section  of  the  rib. 

Again, 

cos  6'  =  cos  {B  —  t)  =  cos  &  -\-i  sin  0 ; 
sin  ff'  =  sin  (^  —  i)  —  sin  8  —  i  cos  0. 


]'■ 

i 

1 
1 

Iv- 

1 

I'li'  ' 

■ 

Ili  '!' 

K  ■ 

Wi  ■ 

i~'^' 

K'.'i 

786 


THEORY  OF  STRUCTURES. 


dx'      dx 


.dy 


dy'       dy        .dx 


/T/  =  -xH-^v7     and     -^,  =  -r-i 


"  ds'~  ds 


ds 


ds 


.      (12) 


Hence,  approximately, 


.    d 


.dy 


d 


Ax 


-rM'-x)  =  i-i-   and   -i:Ly'-y)^-i-j, 


ds 


ds 


ds 


Thus,  if  X  and  Y  are  respectively  the  horizontal  and  verti- 
cal displacements, 

•  dX       .dy         ^    dY  .dx 

-j-=i-T-    and     -J-  =  —  t-r, 
ds         ds  ds  ds 


or 


dX 


=  t 


dY 


(13) 


dy       "  dx 

16.  Effect  of  T  and  of  a  Change  of  r  in  the  Temperature. 

ds'  =  ds[i--^. 

Also,  if  there  is  a  change  from  the  mean  of  t°  in  the  tem- 
perature, the  length  ^'^(i  -  -^  must  be  multiplied  by 
(i  ±  e^,  e  being  the  coefficient  of  linear  expansion. 

,'.  ds' =:  ds[i  -  ^^{\  ±  et) 

~  '•;(  I  -  T^  ±  €^)»  approximately.         (14) 
By  equations  (12), 
dx'  =  {dx-^i.dy)-^~{dx-\-i.dy)[i  -  j^  ±et) 

and  ' 


ds' 


dy'  =  (dy  -  i-dx)-r  =  {dy  -  i.dx)\\ 


i.dx){i--^±ei). 


GENERAL  EQUATIONS. 


787 


.-.  dX  —  d{x'  —  4:)  =  idy  —  f^r-r  =F  ^t\dx 


and 


^F=  d{y'  —y)z=—  idx  —  l^  q:  et^dy^  approxiniately, 

Hence, 

.dy  .  /-V  T 


and 


X  =  ^'  -  .  =  f^i^dx  -  J^'l^  T  e/)rf^    .(15) 
Y=y-y=  -  fidx  -  fil-^  T  e,)|rf..     (,6) 


iVb^f'. — A  nearer  approximation  than  is  given  by  the  pre- 
ceding results  may  be  obtained  as  follows: 

Let  X  -\-  dx,  y  -\-  dy  he  the  co-ordinates  of  a  point  very 
near  C  before  deformation. 

Let  x'  +  dx' ,  y'  +  dy'  be  the  co-ordinates  of  a  point  very 
near  C  after  deformation. 

Then 

ds*  -  dx^ -{- df    and    ds'^  -  dx"  ^  dy'\ 
.-.  ds'^  -  ds'  -  dx"  -  dx-"  -t-  dy"  -  dy'\ 


or 


{ds'-ds){ds'  +  ds)  =  {dx'-dx){dx'  +  dx)  +  {dy'-dy){dy'  +  ^). 
.'.  (i/j'  —  ds)ds  =  {dx'  —  dx)dx  -\-  {dy'  —  dy)dy,  approximately. 


.'.  dx'  -dx  =  {ds'  -  ds)^  -  {dy'  -  dy)$ 


'dx 


dx 


and 


ds  dx       ,  -  ,        ,  .dx 
dy'  -dy  =  {ds'  -  ds)^^  ^  -  (^-^  -  dx)-^y 


Hence,  by  equations  (12)  and  (14), 


dx'  -dx  =  tf^dx  -  -^[^j,]  dx  ±  e/(^)  dx 


I 


''^'i  II      h 


788 
and 


THEORY  OF  STRUCTURES. 


,  ,        ,  ..  T  (dsVdx  ^  IdsV....   , 

df  —  dy  =  —  idx  —  -E-A-nA  —..dx  ±  e/l-7-  J  -j-dx. 


'dx 

\dx  i  dy 


EAvixl  dy 


and 


'  y'  -y  =  -  fj'^''  -S^ea[jx)  ^^-^  ±  ''S^^x)  '■^^•"• 


:/  dy 


These  equations  are  to  be  used  instead  o.  equations  (15) 
and  (16),  the  remainder  of  the  calculations  being  computed 
precisely  as  before. 

The  following  problems  are,  in  the  main,  the  same  as  those 
given  in  Art.  180  of  Rankine's  Civil  Engineering,  13th  edition- 

17.  Rib  of  Uniform  Stiffness. — Let  the  depth  and  sectional 
form  of  the  rib  be  uniform,  and  let  its  breadth  at  each  point 
vary  as  the  secant  of  the  inclination  of  the  tangent  at  the  point 
to  the  horizontal. 

Let  A^,  Ij  be  the  sectional  area  and  moment  of  inertia  at 
the  crown. 

Let  A,/  be  the  sectional  area  and  moment  of  inertia  at  any 
point  C,  Fig.  503 

Then  1 

ds 


A  =  A,  sec  6=  A, 


dx 


(17) 


Also,  since  the  moments  of  inertia  of  similar  figures  vary 
as  the  breadth  and  as  the  cube  of  the  depth,  and  since  the 
depth  in  the  present  case  is  constant, 


/  sec  ^  =  /  — 


•  •     •    • 


.    (I8j 


Again,  -^  =  -5 a  =  X'  ^^^  *^^  intensity  of  the  thrust 

A       A^  sec "      ^, 

is  constant  throughout.  ' 


ARCHED  RIB   OF  UNIFORM  DEPTH.  789 

Hence,  equations  (5),  (15),  and  (16),  respectively,  become 

«  =  ^-;^/'^<=^^; (19) 


t—rdx 
ax 


H 

EA 


X  ±  etx ; 


.     .     (20) 


■      •    ■    y_^=_^'v/.r_(^q:e,](^_^.).      .    (2,) 

Equation  (19)  shows  that  the  deflection  at  each  point  of  the 
rib  is  the  same  as  that  at  corresponding  points  of  a  straight 
horizontal  beam  of  a  uniform  se  \o\\  equal  to  that  of  the  rib 
at  the  crown,  and  acted  upon  by  the  same  bending  moments. 

Ribs  of  uniform  stiffness  are  not  usual  in  practice,  but  the 
formulae  deduced  in  the  present  article  may  be  applied  without 
sensible  error  to  flat  segmental  ribs  of  uniform  section. 

18.  Parabolic  Rib  of  Uniform  Depth  and  Stiffness,  with 
Rolling  Load ;  the  Ends  fixed  in  Direction ;  the  Abut- 
ments immovable. 


Pig.  504. 


Let  the  axis  of  ;r  be  a  tangent  to  the  neutral  curve  at  its 
summit. 

Let  k  be  the  rise  of  the  curve. 

Let  X,  y  be  the  co-ordinates  at  any  point  C  with  respect 
to  0. 

Then 


=  ^^t-')' 


(22) 


and 


\% 


790  THEORY  OF  STRUCTURES. 

Let  w  be  the  dead  load  per  horizontal  unit  of  length. 

<i    jy'  u     (<     live       "       "  "  "      "       " 

Let  the  live  load  cover  a  length  DE,  =  r/,  of  the  span. 
Denote  by  (A)  formulae  relating  to  the  unloaded  division 
OE,  and  by  (B)  formulae  relating  to  the  loaded  division  DE. 
Equations  (7)  and  (8),  respectively,  become 

(A)  S=S,-\-\^j^-w)x; (24) 

(B)  S=S,-^^-i^x-w'\x-{i-r)l\.     .    .    (25) 

^A)    M^M.  +  S.x+l^-w)'^',.    ,     .    .    .    .    .    (26) 

(B)    M  =  M,-{.S,x+[-jr--w)---{x-(i-r)l\\   (27) 

Since  the  ends  are  fixed, 

?;  =  o  =  >\. (28) 

Hence,  by  equations  (19)  and  (26), 

(A)  i=:-^J^M,x^sf^+(^-^^-^)^y,    .     (29) 

and  by  equations  (19)  and  (27), 

I     i  .^      .    e-*'  ,   f8/&^         \x*    ^ 

(B)  ,=  -  — |^„^H-.S.-+(-^-«;J^ 

-^{4r-(i-r)/{'|.     (30) 
When  X  —  l,i=  /,  =  o,  and  therefore,  by  the  last  equation, 

:  .r,,V    o  =  ^  +  ^-+i-75--«'J6-6-'''/'.   .    .    .    (31) 


S5!^  1 


ARCHED   RIB   OF   UNIFORM  DEPTH. 


dv 


791 


Again,  let «  =  -7- .     Then 


*/  a  t/  a 


■dy  ,    _     Hdv  dy      _  .dy 
dx       ~    1     dx  dx      ~   ^dx 


But  t,  =  o,     and     -j^  =  -y 


By  the  conditions  of  the  problem,  x'  —  x  and  ^'  —  ^  are 
each  zero  at  Q.  Hence,  equations  (20)  and  (21),  respectively, 
become 

o=-^/'/'^'^^'- (^^^^)^;  •  •  (33) 

o=-  J^    idx. (34) 


Substitute  in  eqs.  (33)  and  (34)  the  value  of  i  given  by  eq. 
(30),  and  integrate  between  the  limits  o  and  /.     Then 


and 

°  =  -£7,i-2-+-6-+l— -«^)i5-«"'iir 

which  may  be  written 

•  '     "4  '  \  /  ^20  20 


;li       r 


'      '  (1 

•1     i . 

i:    !^ 


792 
and 


THEORY  OF  STRUCTURES,     j 


o=J/.  +  5.^  +  (^-«;)^-z.V:^.  .    .    .     .    .    (36) 

Hence,  by  eqs.  (31),  (35),  (36),  ,  "^  * 


wl       w  l^r^ 

M,  = 

12  3 


(-^) 


4  '3 


(37) 

(38) 


-fT; 


8" 


4       8      '  4  y     4     ^ 


K'+^iiii-) 


.  .  (39) 


When  x  =  l,M=M„  and  5  =  S,, 
Hence,  by  eqs.  (25)  and  (2y), 


and 


S,  =  S,-\-  \^-ji w]l  -  w'rl, 


n^        ,^    .    c,  ,   /8/fe^        Y      w'r'P 


Substituting  in  these  equations  the  values  of  5„,  M^,  given 
above,  we  have 


and 


5,=  -^-»V/(,-^  +  l)+^,    .    .    .    (40) 
J/,  = wTr' r  +  -)  +  -/&^.  .     .     (41) 


To  find  the  greatest  intensity  of  stress,  etc. — The  intensity  of 

.  T      H 

the  stress  due  to  direct  compression  =  -r  =  -^  . 


ARCHED  RIB   OF   UNIFORM  DEPTH. 


793 


The  intensity  of  the  stress  in  the  outside  layers  of  the  rib 
due  to  bending  is  the  same  as  that  in  the  outside  layers  of  a 
horizontal  beam  of  uniform  section  A^  acted  upon  by  the  same 
moments  as  act  on  the  rib,  for  the  deflections  of  the  beam  and 
rib  are  equal  at  every  point  (eq.  (19) ).  Also,  since  the  rib  is 
fixed  at  both  ends,  the  bending  moment  due  to  that  portion  of 
the  load  which  produces  flexure  is  a  maximum  at  the  loaded 
end,  i.e.,  at  Q.     Hence,  the  maximum  intensity  of  stress  (/,) 

•LJ 

occurs  at  Q,  and/*,  =  ^-  ±  M^j,  z^  being  the  distance  of  the 

layers  from  the  neutral  axis. 

H  and  J/,  are  both  functions  of  r,  and  therefore /»i  is  an  ab- 
solute maximum  when 


But 


dr~^~Adr^I,dr ^42> 

dH      iSw'l'   r\i-  rf 


and 


dr 


dM. 


4     k 


1+^ 


45    /.   ' 


4  Ak^ 


Hence, /i  is  an  absolute  maximum  when 

I  I        2  kz\ 


Qz^wTr{\  -rf' 


^(■+f.4=) 


^7; 


(43) 


The  roots  of  this  equation  are 
r  =  \ 

45   /. 


and 


0^+4  A,k' 


2  A^zji 


±  I 


(45) 


!       I  ■ 


I       1^ 


794 


THEORY  OF  STRUCTURES. 


r  =  I  makes  tt  ^^^o,  so  that  the  maximum  value  of  p. 
corresponds  to  one  of  the  remaining  roots. 

■     Thus,  "■'     .■";'    *••■     :  .    ■;    ,  :■• 

Xht  max.  thrust  = -j-\H -{-——■  Ma  =p^'    .     .    (46) 

and 

the  max.  tension  =  -j-\—  ^ -\- ~j~^^-)  —Px't       (47) 
■"1  •'1 

the  values  of  H  and  J/,  being  found  by  substituting  in  eqs. 
(39)  and  (41) 


r  = 


4  A,k' 


5.-i^'- 


2A,2^k 


or 


2    ^  4^,/^' 


1  + 


2  y4,2',/^     . 


(48) 


according  as  the  stress  is  a  thrust  or  a  tension. 

If  eq.  (47)  gives  a  negative  result,  there  is  no  tension  at  any 
point  of  the  rib. 

Note. — The  moment  of  inertia  may  be  expressed  in  the  form 

q  being  a  coefficient  depending  upon  X}\q.  form  of  the  section. 
Hence, 

the  maximum  intensity  of  stress  =  -j\±.  H -\ ' j .     .    (49) 

Corollary  i. — If  the  depth  of  the  rib  is  small  as  compared 
with  k,  the  fraction  t  will  be  a  small  quantity,  and  the  maxi- 
mum intensity  of  stress  will  approximately  correspond  to  r  =  |^. 


ARCHED  RIB  OF   UNIFORM  STIFFNESS. 


;95 


The  denominator  in  eq.  (39)  may  be  taken  to  be  k,  and  it  may 
be  easily  shown  that  the  values  of/,',//'  are 

^•-^.  (    8  U+T;fe'J^4-^-+^5-^|:-    (50) 

Cor.  2.  If   the   numerator  in  eqs.  (48)  is  greater  than  the 
denominator,  then  r  must  be  unity.     Hence,  by  eq.  (39)  and 

_  r  zv  -\-  w'      /[S^eiEI^ 

8       bk      ^'^~blF'' (52) 

and  by  eqs.  (38)  and  (41), 

M       M       ^'  (      I       /^  I  -  ^  ,   1 5  ^fET, 


-     16'      b     ^k^^\'b~T 


(53) 


Thus, //,//' can  be  found  by  substituting  these  values  of 
//"and  J/,  in  eqs.  (46)  and  (47). 

19.  Parabolic  Rib  of  Uniform  Stiffness,  hinged  at  the 
Ends. 

Let  the  rib  be  similar  to  that  of  the  preceding  article. 

Since  the  ends  are  hinged,  M^  —  O  —  M^,  while  i  is  an  un- 
determined constant. 

The  following  equations  apply : 


(A)  i=:i;+^___^J^. ^j^^ 

(B)  S=iS,-\-\-j^  -w)x-w'\x-{\-r)l\',      .     (55) 


796 


THEORY  OF  STRUCTURES, 


<A)  M=S,x^\^^-w]\; (56) 

<B)  iJ/=5,;r+(^-«')f-^'l^-(i-r)/r.        (57) 

.       .         I    i  ^;r"       l%kH         \x^ ) 

*  = '•  -  £/i -^"T  +  IT'""  -  H  6  -  6  ^'^-('-'')^^   J  •  (59) 

Assume  that  the  horizontal  and  vertical  displacements  of 
the  loaded  end  are  ni/. 

Substitute  in  eqs.  (20)  and  (21)  the  value  of  i  given  by  eq. 
(59).  Integrate  and  reduce,  neglecting  the  term  involving  the 
temperature.     Then 


0  =  ta 


^{-^"Fi+V-T^-^ 


I 


UH 


From  (57),  since  J/,  =  o, 

o  =  5„  +  \^-jr  -  wj-  -  w'l-.    ,    .    .    (62) 

Equations  (60),  (61),  and  (62)  are  the  equations  of  condi- 
tion. 

Subtract  (61)  from  (60).     Then 

which  may  be  written 

^    ,   (SkH         W  ,Jr*      r'\  „r   I      ,^  ^ 


5 


ARCHED  RIB   OF   UNIFORM  STIFFNESS. 

Subtract  {61)  from  (62).     Then 


m 


(UH 


Hence, 


kH        v/  ,  [r*      r*      r'\  r    1 


H  = 


^l^  +  -2(Sr'-Sr*-\-2r')^ 


•    •     (65) 


V    ^  8  A.J^'J 
Eliminating  S,  between  (61)  and  (62), 

^»  =  -j7l(T--)^^--'Kr2-2-4)}-    (66) 

Also,  by  (55), 

^  _  „    ,   (8^1/        \ 

•^i  —  '^o-t-  \-jr  —  IV jl  -  w'rl—  —  p^  suppose.     {6^) 

Eliminating  5„  between  (62)  and  {^fj, 

-^=S,  =  [-j^-w]^-w'/{r-~).      .     .     .     (68) 


Eqs.  (62),  (65),  (66),  and  (68)  give  the  values  ol  If  S    S 

and  ?■„.  '     0'    1, 

Again,  the   maximum    bending  moment   Af'  occurs   at  a 
pomt  given  by  ~  =  o  in  (57),  i.e., 


o  -  -^0  +  [-J. wjx  -  w'{x  -  (i  -  ry\. 

Subtract  (69)  from  {6y).     Then 

'  _       (Skff         \ 

-Px  =  S,  =  \~^-w){l~x)-wXl-jc). 


•    (69) 


fff  i 


798 


THEORY  OF  STRUCTURES. 


Hence,  the  distance  from  the  loaded  end  of  the  point  at 
which  the  bending  moment  is  greatest  is 


/-;ir  = 


w  -f-  ^v' 


UH' 


•         •        • 


.    (70) 


Substitute  this  value  of  x  in  (57),  and,  for  convenience,  put 


,      UH 
w  -\-w -pr  =  '«• 


Then 


^'  =  ^.['-9+—.'" 


\         ;«/  2  \      w    '       / 


.  /  „       w'  —  m        w'     \       P 
=  ^i"^»  +  —7-^  -  7 '■'^j  -  ;«  l-^»+  {^'-fn)l-  w'rl\ 


_,P^l'^'  —  i>t       w'\ 


m 


But  by  (62),  0  =  5.  +  ^L^i  _  '^f.Vv, 


^  '       vr         '   '   m\      2  / 


2m 


Hence,  J/',  the  maximum  bending  moment, 

P* 


~    /      ,      ,       ZkH\ 

2\W  -\-W jj— J 

As  before,  the  greatest  stress  (a  thrust) 


(71) 


•    .    •    (7? 


and  the  value  of  r  which  makes  //  an  absoiu,     m?   imum  is 

dp' 
given  by  -^  =  o.     But  by  (71),  M'  involves  r"  in  the  numera- 


ARCHED  KIB  OF  UNIFORM  STIFFNESS. 


799 


dp' 

tor  and  r"  in   tlic    denomhutor,  so  that  -,—  =  O  will   be  an 

dr 

equation  involving  r'\ 

One  of  its  roots  isr  =  i,  which  generally  gives  a  niimuiinn 
value  of  /»,'.  Dividing  by  r  —  l,  tlie  equation  reduces  to  one 
of  \\m  thirteenth  order,  but  is  still  far  too  complex  for  use.  It 
is  found,  however,  that  ^  =  i  gives  a  close  approximation  \.<o  the 
absolute  maximum  thrust. 

With  this  value  of  r,  and,  for  convenience,  putting 


,    15  /,    I 


By  (65). 
By  (62), 

By  (68), 
By  (66), 
By  (70), 


'  2  (  \     '    2  /    «      '4 


l-X 


[w 


2t>  \n 


« 


I         7V 


By  (71), 


M'= 


,  sU^^+tJ— -+4f 


w  \n 
7U  4-  - 


—  I       w 
n        '2 


(74) 

(75) 
(76) 

(77) 


(78) 


iV(?/t'. — If  the  rib  is  merely  supported  at  the  ends  but  not 
fixed,  the  horizontal  displacement  of  the  loaded  end  may  be 


8oo 


m:>^:%:    THEORY  OF  STRUCTURES. 


represented  by  jaH  {Art.  ii).  Thus  the  term  —  fiH  must  be 
added  to  the  right-hand  side  of  eq,  (15). 

20.  Parabolic  Rib  of  Uniform  Stiffness,  hinged  at  the 
Crown  and  also  at  the  Ends. — In  this  case  M  —  o  dX  the 
crown,  which  introduces  s.  fourth  equation  of  condition. 

By  (57). 

which  may  be  written 

o  =  S,^^-^-w)^-w'l[r^-r+'^.    .    .    (79) 
Eliminating  5,  between  (79)  and  (62), 


Hence, 

By  (79)» 
By  (68), 
By  (66), 


— — w  =  w  {—  2r'  -\- <\r  —  i). 


ff  =  :^^\w  -  w\2r' -  4r  +  i)\.     .    .     .    (80) 


P^S^=^^r-.r 


Zr.    = 


wT 

24EI, 


By  (/o)  and  (82), 


'-X=: 


w'l,  ., 


(81) 
(82) 


(1-4^  +  4^'-/) (83) 


iw'ir  —  \f      4 


=  "••••••••    v"4) 


By  (71). 


w'l 


M'  =  — g-(r  -•-  i)' ,    (85) 


JILMJ-M...  I'* 1 


«-WSKS^ 


PARABOLIC  RIB   OF   UNIFORM  STIFFNESS. 

When  r  =  i 


/7 

—    n 


w' 


^=8i'r+-J.  s.= 


2 

r  li 


8  '  "^^  ~   8  ' 


'•  -  -  384  i:7. ' 


'/» 


and 


J/'  = 


64- 


801 


(86) 


These  results  agree  with  those  of  (73)  to  (78),  if  «  =  I. 
Lt  general,  when  n  =■  i, 


4) 


!5) 


w  +  —  (5r'  —  5/  +  2O  =  w  —  w'(2r'  —  4r  -f- 1), 

by  (65)  and  (80).     Htnce, 

2r'  —  5r*  +  9^'  +  Sr  +  2  =  o  =  {2r  —  i)(r  -  i)'(r'  —  2), 

and  the  roots  are  r  =^^,  r  —  \,  r  ^=  ±  1^2. 

Hence,  ?i  =  l  only  renders  the  expressions  in  {'66)  identical 
with  the  corresponding  expressions  of  the  preceding  article 
when  ;/  =  ^  or  I. 

Again,  the  intensity  of  thrust  is  greatest  at  the  outer  flange 
of  the  loaded  and  the  inner  flange  of  the  unloaded  half  of  the 
rib,  and  is  . 


_     /'     )Z,W' 


\l 


SA,   U,  S    '   k 


\w^ 


-)!• 


The  intensity  of  tension  is  greatest  at  the  inner  flange  of  the 
loaded  and  the  outer  flange  of  the  unloadv;d  half  of  the  rib, 
and  is 


/'      (  ^'i  If/'  I  /         ,    w' 


8^,   (/, 


The  greatest  total  horizontal  thrust  occurs  when  r  =  i,  and 
its  value  is 


U 


(w  -f-  w'). 


802 


THEORY  OF  STRUCTURES. 


the  deflection  is  an  absolute  maximum  when    "-\y'  —  y) '=  o. 


21.  Maximum  Deflection  of  an  Arched  Rib. — The  deflec- 
tion must  necessarily  be  a  maximum  at  a  point  given  by  i  =:^  o. 
Solve  for  x  and  substitute  in  (i6)  to  find  the  deflection  j/'  —  y; 

d 

dr' 

The  resulting  equation  involves  r  to  a  high  power,  and  is  too 
intricate  to  be  of  use.  It  has  been  found  by  trial,  however, 
that  in  all  ordinary  cases  the  absolute  maximum  deflection 
occurs  at  the  middle  of  the  rib,  when  the  live  load  covers  its 

/ 
whole  length,  i.e.,  when  x  =  -,  and  r  =  i. 

Case  I.  Rib  of  An.  i8.    For  convenience,  put  i  +  —  Vra= J. 

^         '    4  Ak 

Then,  by  (39), 


(87) 


By  (38)  and  (41), 


-M.=  -{w  +  wi-—^:^l'-l^  =  -M,.,    (88) 


12 


4  J     k 


By  (36)  and  (38), 

By  (30),  (38),  (89),  ■. 


(89) 


-  ^^{^M,x -  zM^ -{- 2M^.    .    .    .     (90) 


Ei: 

Hence,  the  maximum  deflection 

-      J,  tdx  =  -  -I    \x-i-j^  2j,)dx  =  -  ^-^--     - 

r     W  -\-  W'  S  —  I  5      6//'  ^ 


MAXIMUM  DEFLECTION   OF  AN  ARCHED  RIB.  803 

The  central  deflection  d^  of  a  uniform  straight  horizontal 
beam  of  the  same  span,  of  the  same  section  as  the  rib  at  the 
crown,  and  with  its  ends  fixed,  is 

'^'~ iH~Ejr'  ••••••  (92) 

Hence,  neglecting  the  term  involving  the  temperature, 

^■  =  ^^" (93) 

Case  II.  Rib  of  Art.  19. 
By  (65), 

r-  _    l^  W  -\- W' 

u~~Ji (94) 

•    By  (66)  and  (62), 

^»-2-i^/>  +  ^)-7r  =Ill7-    •    •    •    (95) 
By  (30),  (94),  and  (95), 

EI,\i2       2  "^3/7 (96) 

Hence,  the  maximunr.  deflection 

If  the  ends  of  th    Seam  in  Case  I  are  free,  its  central  de- 
flection 

=    5   ^^±^1_,, 
384       £f      ~    '  ' 

•■•^■'  =  ^'^'' (98) 

Thus,  the  deflection  of  the  arched  rib  in  both  cases  is  less 
than  that  of  the  beam. 


8o4 


THEORY  OF  STRUCTURES. 


22.  Arched  Rib  of  Uniform  Stiffness  fixed  at  the  Ends 
and  connected  at  the  Crown  with  a  Horizontal  Distribut- 
ing Girder. — The  load  is  transmitted  to  the  rib  by  vertical 
struts  so  that  the  vertical  displacements  of  corresponding 
points  of  the  rib  and  girder  are  the  same.  The  horizontal 
thrust  in  the  loaded  is  not  necessarily  equal  to  that  in  the  un- 
loaded division  of  the  rib,  but  the  excess  of  the  thrust  in  the 
loaded  division  vvill  be  borne  by  the  distributing  girder,  if  che 
rib  and  girder  are  connected  in  such  a  manner  tiiat  the  hori- 
zontal displacement  of  each  at  the  crown  is  the  same. 

The  formulae  of  Art.  i8  are  applicable  in  the  present  case 
with  the  modification  that  /,  is  to  include  the  moment  of 
inertia  of  the  girder. 

The  maximum  thrust  and  tension  in  the  rib  are  given  by 
equations  (64)  and  (65). 

Let  z'  be  the  depth  of  the  girder,  A'  its  sectional  area. 

H  M  z' 

The  greatest  thrust  in  the  girder  =  — — j — —  -j — ~.    (99) 


The  greatest  tension  in  the  girder  = 


A,^A'   '   2EI, 
2EL 


.  (100) 


H  and  J/,  being  given  by  equations  (66)  and  {6y),  respectively. 

The  girder  must  have  its  ends  so  supported  as  to  be  capable 
of  transmitting  a  thrust. 

23.  Stresses  in  Spandril  Posts  and  Diagonals. — Fig.  505 
represents  an  arch  in  which  the  spandril  consists  of  a  series  of 
vertical  posts  and  diagonal  braces. 


Fig.  505. 


Let  the  axis  of  the  curved  rib  be  a  parabola.  The  arch  is 
then  equilibrated  under  a  uniformly  distributed  load,  and  the 
diagonals  will  be  only  called  into  play  under  a  passing  load. 


STJ^£:SS£S  IN  SPANDRIL   POSTS  AND  DIAGONALS.     805 

Let  X,  y  be  the  co-ordinates  of  any  point  F  of  the  parabola 
with  respect  to  the  vertex  C.     Then 

y  =  V^  • 

Let  the  tangent  at  F  meet  C£  in  L,  and  the  horizontal  ££ 
in  G. 

hGt  BC=k'.     Then 

BL  =  BC  -  CL  =  BC  -  CN  =  k'  -  y. 

J^et  A^  be  the  total  number  of  panels. 

Consider  any  diagonal  FB  between  the  nth  and  {n  -\-  i)th 
posts. 

Let  za'  be  the  greatest  panel  live  load. 

The  greatest  compression  in  £D  occurs  '^•Lcn  the  passing 
load  is  concentrated  at  the  first  ?i  —  i  panel  points. 

Imagine  a  vertical  section  a  little  on  the  left  of  £F. 

The  portion  of  the  frame  on  the  right  of  this  section  is 
kept  in  equilibrium  by  the  reaction  R  at  P,  and  by  the  stresses 
in  the  three  members  met  by  the  secant  plane. 

Taking  moments  about  G, 

D.GF  cose  =  R.  AG, 

D  being  the  stress  in  DE,  and  B  the  angle  DEP, 
Now, 


R  = 


w  n\ 


•) 


is 
le 


Also, 


X -^  GB _k'  -\- y  ^ 
GB     ~  k'  -y'' 


GB 


_  k'x  —  xy 

-  ^7~' 


and  hence, 

GE=.GB^x^^^-^^^^^-,    and 

2y 

Hence, 

_  w'  n{n  —  i)ly  -\-  k'x 


t       k'x  — 
GA  =-  +  1^ — 
2  2y 


xy 


D-  — 
2 


xy 


N 


k'x  -(-  xy 


sec  B. 


The  stresses  in  the  counter-braces  (shown  by  dotted  lines  in 
the  figure)  may  be  obtained  in  the  same  manner. 


8o6 


THEORY  OF  STRUCTURES. 


I  ■  The  greatest  thrust  in  EF  =  w'  -|-  w. 

The  greatest  tension  in  EF  ~  Bcos  0  —  w,  w  being  the 
dead  load  upon  EF. 

If  the  last  expression  is  negative,  EF  is  never  in  tension. 

24.  Clerk  Maxwell's  Method  of  determining  the  Re- 
sultant Thrusts  at  the  Supports  of  a  Framed  Arch. — Let 
As  be  the  change  in  the  length  s  of  any  member  of  the  frame 
under  the  action  of  a  force  P,  and  let  ci  be  the  sectional  area  of 
the  member.     Then 

±  -jT-s  =  As, 
.  Ea 

the  sign  depending  upon  the  character  of  the  stress. 

Assume  that  all  the  members  except  the  one  under  con- 
sideration are  perfectly  rigid,  and  let  A/  be  the  alteration  in 

the  span  /  corresponding  to  As.      The  ratio  --  is  equal  to  a 

constant  m,  which  depends  only  upon  the  geometrical  form  of 
the  frame. 


.-.  A/  =  in.  As  =  ±  mP-fT- . 

ha 

Again,  P  may  be  supposed  to  consist  of  two  parts,  viz.,/, 
due  to  a  horizontal  force  H  between  the  springings,  and  /,  due 
to  a  vertical  force  F  applied  at  one  springing,  while  the  other 
is  firmly  secured  to  keep  the  frame  from  turning. 

By  the  principle  of  virtual  velocities, 


Al 


A 


Similarly,  ~  is  equal  to  some  constant  n,  which  depends 
only  upon  the  form  of  the  frame. 


.'.  Al=  ±  {m'H ^  mnV)-^. 


ds 


CLERK  MAXWELL'S  METHOD.  807 

Hence,  the  total  change  in  /  for  all  the  members  is 

If  the  abutments  yield,   let  '2,Al  =  ^J^,  /<  being  some  co- 
efficient to  be  determined  by  experiment.     Then 


/r= 


±  2 [mn V- 


s 
Ea 


If  the  abutments  are  immovable,  2AI  is  zero,  and 
ri  =■  — 


(C) 


<''4J 


.     .    (D) 


Fis  the  same  as  the  corresponding  reaction  at  the  end  of  a 
girder  of  the  same  span  and  similarly  loaded.  The  required 
thrust  is  the  resultant  of  H  and  V,  and  the  stress  in  each 
member  may  be  computed  graphically  or  by  the  method  of 
moments.     In  any  particular  case  proceed  as  follows  : 

(1)  Prepare  tables  of  the  values  of  m  and  n  for  each  member. 

(2)  Assume  a  cross-section  for  each  member,  based  on  a 
probable  assumed  value  for  the  resultant  of  V  and  H. 

(3)  Prepare  a  table  of  the  value  of  Pi^-pr  for  each  member, 


and  form  the  sum  2 


('"■iJ- 


(4)  Determine,  separately,  the  horizontal  thrust  between 
the  springings  due  to  the  loads  at  the  different  joints.  Thus, 
let  v^ ,  7',  be  the  vertical  reactions  at  the  right  and  left  supports 

due  to  any  one  of  these  loads.     Form  the  sum  ^y-ninV-jr-L 

using  v^  for  all  the  members  on  the  right  of  the  load  and  7',  for 
all  those  on  its  left.     The  corresponding  thrust  may  then  be 


111 
I 


8o8 


THEORY  OF  STRUCTURES. 


found  by  eq.  (C)  or  eq.  (D),  and  the  total  thrust  H  is  the  sum 
of  tlie  thrusts  due  to  all  the  weights  taken  separately. 

(5)  Repeat  the  process  for  each  combination  of  live  and 
dead  load  so  as  to  find  the  maximum  stresses  to  which  any 
member  may  be  subjected. 

(6)  If  the  assumed  cross-sections  are  not  suited  to  thes^ 
maximum  stresses,  make  fresh  assumptions  and  repeat  the 
whole  calculation. 

The  same  method  may  be  applied  to  determine  the  result- 
ant tensions  at  the  supports  of  a  framed  suspension-bridge. 


Note. — The  formulae  for  a  parabolic  rib  may  be  applied 
without  material  error  to  a  rib  in  the  form  of  a  segment  of  a 
circle.  More  exact  formulae  may  be  obtained  for  the  latter  in 
a  manner  precisely  similar  to  that  described  in  Arts.  18-22, 
but  the  integrations  will  be  much  simplified  by  using  polar  co- 
ordinates, the  centre  of  the  circle  being  the  pole. 


EXAMPLES. 


809 


EXAMPLES. 

1.  Assuming  that  an  arch  may  be  divided  into  elementary  portions 
by  imaginary  joint  planes  parallel  to  the  direction  of  the  load  upon  the 
arch,  find  the  limiting  span  of  an  arch  with  a  horizontal  upper  surface 
and  a  parabolic  soffit  (latus  rectum  =  40  ft.;,  the  deptii  over  the  crown 
being  6  ft.  and  the  specific  weight  of  the  load  120  lbs.  per  cubic  foot; 
the  thrust  at  the  crown  is  horizontal  (:::;  P)  and  4  ft.  above  the  soffit. 

2.  A  masonry  arch  of  90  ft.  span  and  30  ft.  rise,  with  a  parabolic  in- 
trados  and  a  horizontal  extrados,  springs  from  abutments  with  vertical 
faces  and  10  ft.  thick,  the  outside  faces  being  carried  up  to  meet  the 
extrados.  The  depth  of  the  keystone  is  3  ft.  The  centre  of  resistance 
at  the  springing  is  the  middle  of  the  joint,  and  at  the  crown  12  in.  below 
the  extrados.  Tlie  specific  weight  of  the  masonry  may  be  taken  at  150 
lbs.  per  cubic  foot.  Determine  (a)  the  resultant  pressure  in  the  vertical 
joint  at  the  crown  ;  (p)  the  resultant  pressure  in  the  horizontal  joint  at 
the  springing  ;  (c)  the  maximum  stress  in  the  vertical  joint  aligning  with 
the  inside  of  an  abutment. 

3.  The  intrados  of  an  arch  of  100  ft.  span  and  20  ft.  rise  is  the  segment 
of  a  circle.  Tlic  arch  ring  has  a  uniform  thickncs-^  of  3  ft.  and  weighs 
140  lbs.  per  cubic  foot ;  the  superincumbent  load  may  be  taken  at  480 
lbs.  per  lineal  fcjot  of  the  ring.  Determine  the  mutual  pressures  at  the 
key  and  springing,  their  points  of  application  being  2  ft.  and  i^  ft.,  re- 
spectively, from  the  intrados.  Also  find  the  curve  of  the  centres  of  pres- 
sure. 

4.  The  soffit  of  an  arch  of  30  ft.  span  and  10  ft.  rise  is  a  transformed 
catenary.  The  masonry  rises  10  ft.  over  the  crown,  and  the  specific 
weight  of  the  load  upon  the  arch  may  be  taken  at  120  lbs.  per  cubic  foot. 
Determine  the  direction  and  amount  of  the  thrust  at  the  springing. 

5.  A  concrete  arch  has  a  clear  spring  of  75  ft.  and  a  rise  of  7A  ft. ;  the 
heigiit  of  masonry  over  crown  =  5  ft. ;  the  weight  of  the  concrete  =  144 
lbs.  per  cubic  foot.  Determine  the  transformed  catenary,  the  amount 
and  direction  of  the  thrust  at  the  springing,  and  the  curvatures  at  the 
crown  and  springing. 

Alls.  ;;/  =  23.9;  thrust  =  91,354  lbs. ;  slope  at  springing  =  25!° ; 
radius  of  curvature  =  114.2  ft.  at  crown  and  =248.7  ft.  at 
springing. 

6.  Determine  the  transformed  catenary  for  an  arch  of  60  ft.  span  and 
15  ft.  rise,  the  masonry  rising  6  ft.  over  the  crown  and  weighing  120  lbs. 
per  cubic  foot.  Also  find  the  amount  and  direction  of  the  thrust  at  the 
abutments. 


8io 


THEORY  OF  STRUCTURES. 


7.  Determine  the  transformed  catenary  for  an  arch  of  30  ft.  span  and 
']\  ft.  rise,  the  height  of  masonry  over  the  crown  being  \\  ft.  ;  weight  of 
the  masonry  =  125  lbs.  per  cubic  foot.  Also  find  the  thrust  at  the  spring- 
ing and  the  curvature  at  the  crown  and  the  springing. 

8.  In  a  parabolic  arch  of  50  ft.  span  and  10  ft.  rise,  hinged  at  both 
ends,  a  weigiit  of  i  ton  is  concentrated  at  a  point  whose  horizontal  dis- 
tance from  the  crown  is  10  ft.  Find  the  total  thrust  along  the  axis  of 
the  rib  on  each  side  of  the  given  point,  allowing  for  a  change  of  60  from 
the  mean  temperature  (e  =  .0000694). 

9.  A  parabolic  arched  rib  of  100  ft.  span  and  20  ft.  rise  is  fixed  at  the 
springings.  The  uniformly  distributed  load  upon  one-half  of  the  arch 
is  100  tons,  and  upon  the  other  200  tons.  Find  the  bending  moment 
and  shearing  force  at  25  ft.  from  each  end. 

10.  An  arched  rib  with  parabolic  axis,  of  100  ft.  span  and  i2i  ft.  rise, 
is  loaded  with  i  ton  at  the  centre  and  i  ton  at  20  ft.  from  the  centre, 
measured  horizontally.  Determine  the  thrusts  and  shears  along  the  rib 
at  the  latter  point,  and  show  how  they  will  be  affected  by  a  change  of 
100°  F.  from  the  mean;  the  coefficient  of  linear  expansion  being  .00125 
for  180°  F. 

11.  A  parabolic  arched  rib  hinged  at  the  ends,  of  64  ft.  span  and  16 
ft.  rise,  is  loaded  with  i  ton  at  each  of  the  points  of  division  of  eight 
equal  horizontal  divisions.  Find  the  horizontal  thrust  on  the  rib,  allowing 
for  a  change  of  60°  F.  from  the  mean  temperature.  Also  find  the  maxi- 
mum flange  stresses,  the  rib  being  of  d(nible-tee  section  and  12  in.  deep 
throughout.     (Coefficient  of  linear  expansion  per  1°  F.  =  /-^  144000.) 

12.  The  axis  of  an  arched  rib  of  50  ft.  span,  10  ft.  rise,  and  hinged  at 
both  ends  is  a  parabola.  Draw  the  linear  arch  when  the  rib  is  loaded 
with  two  weights  each  equal  to  2  tons  concentrated  at  two  points  10  ft. 
from  the  centre  of  the  span.  If  the  rib  is  of  double-tee  section  and  24 
in.  deep,  find  the  maximum  flange  stresses. 

If  the  arch  is  loaded  so  as  to  produce  a  stress  of  10,000  lbs.  per  square 
inch  in  the  metal,  show  that  the  rib  will  deflect  .029  ft.,  E  being  25,000,000 
lbs. 

13.  A  steel  parabolic  arched  rib  of  50  ft.  span  and  10  ft.  rise  is  hinged 
at  both  ends  and  loaded  at  the  centre  with  a  weight  of  12  tons.  Find 
the  horizontal  thrust  on  the  rib  when  the  temperature  varies  60°  F.  from 
the  mean,  and  also  find  the  maximum  flange  stresses,  the  rib  being  of 
double-tee  section  and  12  in.  deep. 

14.  A  semicircular  rib,  pivoted  at  the  crown  and  springings,  is  loaded 
uniformly  per  horizontal  unit  of  length.  Determine  the  position  and 
magnitude  of  the  maximum  bending  moments,  and  show  that  the  hori- 
zontal thrust  on  the  rib  is  one-fourth  of  the  total  load. 

15.  Draw  the  linear  arch  for  a  semicircular  rib  of  uniform  section 


EXAMPLES. 


8ii 


under  a  load  uniformly  distributed  per  horizontal  unit  of  length  (<;)when 
hin,<,'cd  at  botii  ends;  (h)  when  hinged  at  both  ends  and  at  the  centre;  (<) 
when  tixed  at  both  ends. 

!6.  A  semi-elliptic  rib  (axes  2rt  and  2/')  is  pivoted  at  the  springings. 
Find  tlie  position  and  magnitude  of  the  maximum  bending  moment,  the 
load  being  uniformly  distributed  per  iiorizontal  unit  of  length. 

How  will  the  result  be  alTected  if  the  nb  is  also  pivoted  at  the  crown  ? 

17.  Draw  the  equilibrium  polj'gon  for  a  parabolic  arch  of  100  ft. 
span  and  20  ft.  rise  when  loaded  with  weights  of  3,  2,  4,  and  2  tons,  re- 
spectively, at  the  end  of  the  third,  sixth,  eighth,  and  ninth  division  from 
the  left  support,  of  ten  equal  horizontal  divisions.  (Neglect  the  weight 
of  the  rib.) 

If  the  rib  consist  of  a  web  and  of  two  flanges  2^  ft.  from  centre  to 
centre,  determine  the  maximum  flange  stress. 

18.  Find  the  flange  stresses  at  the  ends  of  the  rib,  in  the  preceding 
question,  and  also  at  the  points  at  which  the  weights  are  concentrated, 
when  both  ends  are  absolutely  fixed. 

19.  A  semicircular  rib  of  28  ft.  span  carries  a  weight  of  \  ton  at  4  ft. 
(measured  horizontally)  from  the  centre.  Find  the  tlinist  and  shear  at 
the  centre  of  the  rib  and  at  the  point  at  which  the  weiglit  is  concen- 
trated. 

20.  The  axis  of  an  arched  rib  hinged  at  both  ends,  for  a  span  of  50  ft. 
and  a  rise  of  10  ft.,  is  a  parabola.  Draw  the  equilibrium  polygon  when 
the  arch  is  loaded  with  two  equal  weiglus  of  2  tons  concentrated  at  two 
points  10  ft.  from  the  centre  of  the  span.  Also  flctermine  the  maximum 
flange  stress  in  the  rib,  which  is  a  double-tee  section  2  ft.  deep. 

21.  The  load  upon  a  parabolic  rib  of  50  ft.  span  and  15  ft.  rise,  hinged 
at  both  ends,  consists  of  weights  of  i,  2,  and  3  tons  at  points  15,  25,  and 
40  ft.,  respectively,  from  one  end.  Find  the  axial  thrusts  and  the  shears 
at  these  points. 

Ans.   Horizontal  thrust  =  9.6  tons. 

Axial  thrusts :  above  i  ton  =  9.3  tons  ; 
below  I    "     =97     " 
above  3  tons  =  8.3     " 
below  3    "     =  lo.i    " 

Shears  :  above  i  ton  =  3.  i  tons  ; 

below  I  "  =2.2  " 
above  3  tons  =  5  " 
below  3    "     =  2.6     " 

22.  Draw  the  linear  arch  and  determine  the  maximum  flange  stresses 
for  an  arched  rib  of  80  ft.  span,  16  ft.  rise,  and  loaded  with  live  weights 
each  of  2  tons  at  the  end  of  the  first,  second,  third,  fourth,  and  fifth 
division,  of  eight  equal  horizontal  divisions.     The  rib  is  of  double-tee 


8l2 


THEORY  OF  STRUCTURES. 


section  and  30  in.  deep.  Also  find  the  shears  and  the  axial  thrusts  at 
the  fifth  point  of  division. 

23.  A  wrought-iron  parabolic  iib  of  96  ft.  span  and  i6  ft.  rise  is 
hinged  at  the  two  abutments;  it  is  of  a  double-tee  section  uniform 
throughout,  and  24  in.  deep  from  centre  to  centre  of  tlie  llange?.  Ueter- 
minc  the  compression  at  the  centre,  and  also  the  position  and  amount 
of  the  maximum  bending  moment  {a)  when  a  load  of  48  tons  is  concen- 
trated at  the  centre  ;  {b)  when  a  load  of  96  tons  is  uniformly  distributed 
per  horizontal  unit  of  length. 

Determine  the  deflection  of  the  rib  in  each  case. 

24.  Design  a  parabolic  arched  rib  ol  100  ft.  span  and  20  it.  rise,  hinged 
at  both  ends  and  at  the  middle  joint ;  dead  load  =40  tons  uniformly 
distributed  per  horizontal  unit  of  length,  and  live  load  —  i  ton  per  hori- 
zontal foot. 

25.  Show  how  the  calculations  in  the  preceding  question  are  affected 
when  both  ends  are  absolutely  fixed. 

26.  In  the  framed  arch  represented  by  the  figure,  the  span  is  120  ft., 

the  rise  12  ft.,  the  depth  of  the  truss  at  the 
crown  5  ft.,  the  fixed  load  at  each  top  joint 

Fig.  506.  10  tons,  and  the  moving  load  10  tons.  De- 

termine the  ma.\imum  stress  in  each  member  with  any  distribution  of 
load.  Show  that,  approximately,  the  amount  of  metal  required  for  the 
arch:  the  amount  required  for  a  bowstring  lattice-girder  of  the  same 
span  and  17  ft.  deep  at  the  centre  :  the  amount  required  for  a  girder  of 
the  same  span  and  12  ft.  deep  ::  100  :  155  :  175. 

27.  The  steel  parabolic  ribs  for  one  of  the  Harlem  River  bridges  has  a 
clear  opening  of  510  ft.,  a  rise  of  90  ft.,  a  depth  of  13  ft.,  and  are  spaced 
14  ft.  centre  to  centre.  The  dead  weight  per  lineal  foot  is  estimated  at 
33,000  lbs.  and  the  live  load  at  8000  lbs.  ;  a  variation  in  temperature  of 
75"  F.  from  the  mean  is  also  to  be  allowed  for.  Determine  the  maxi- 
mum bending  moment  (assuming  /constant),  and  the  maximum  deflec- 
tion.    E  =  26,000,000  lbs.     Show  how  to  deduce  the  play  at  the  hinges. 

28.  A  cast-iron  arch  (see  figure)  whose  cross-sections  are  rectangular 
Do'o'    -^  and  uniformly  3  in.  wide,  has  a  straiglit  horizon- 
tal extrados,  and  is  hinged  at  the  centre  and  at 
the  abutments.     Calculate  the  normal  intensity  of 
stress  at  the  top  and  bf)ttom  edges  D,  E  of  the 

Fig.  507.  vertical  section,  distant  5  ft.  from  the  centre  of 

the  span,  due  to  a  vertical  load  of  20  tons  concentrated  at  a  point  dis- 
tant 5  ft.  4  in.  horizontally  from  B.  Also  find  the  maximum  intensity 
of  the  shearing  stress  on  the  same  section,  and  state  the  point  at  which 
it  occurs.     {AB  =  21  ft.  4  in.). 


INDEX. 


Allowance  for  the  weight  of  a  beam,  403. 
Allernaiing  stresses,  152. 
American  iron  columns,  53a. 
Anchorage,  704. 
Angle  of  repose,  237. 
"      "  torsion,  568. 
Angular  momentum,  177, 
Anti-friction  curve,  320. 
"  pivots,  320. 

Arch,  470. 
Arch  abutment,  maximum  thickness  of, 

Arch,  conditions  of  equilibrium  of,  745. 
''      formulx  for  thickness  of,  750. 
"      linear,  743,  750,  760. 

Arched  ribs,  740,  762. 

"         "     deflection  of,  780,  802. 

Arched  ribs,  effect  of   change   of   tem- 
perature oil,  770,  780. 

Arched  ribs,  elementary  deformation  of, 
751. 

Arched  ribs,  general  equations  of  equi- 
librium of,  784. 

Arched  ribs,  graphical  determination  of 
stresses  in,  677. 

Arched    rib   of   uniform  stiffness,    7S8, 
-aij.  795.  800,  804. 

Arched  ribs  with  fixed  ends,  771. 

"      with  hinged  ends,  764. 

Arched  ribs  with  axis  in  form  of  circu- 
lar arc,  769,  773. 

Arched   ribs  with    parabolic   axis,  760, 

775. 
Arched  ribs  with  semicircular  axis,  765, 

775- 
Arches,  middle-third  theory  of,  746. 
Auxiliary  truss,  719. 

Back-stays,  i6,  704. 

Baker's  formulee  for  Strength  of  pillars, 

549- 
Balancing,  I98. 


Beam  acted  upon  by  oblique  forces,  396. 
Beam,  transverse  strength  of,  340,  429. 
Beam,  transverse  vibration    of   loaded, 

461. 
Beams,  equilibrium  of,  93. 

"        of  uniform  strength,  358-365. 
Bearing  surface,  314,  315. 
Belts,  321. 

"      effect  of  high  speed  in,  325. 

"      effective  tension  of,  324. 

"      slip  of,  326. 

"      stiffness  of,  327. 
Bending  moment,  96,  ri8,  434. 
Bending  moment  In  plane  which  is  not 

a  principal  plane,  354. 
Bending  moment,  relation  between,  and 

shearing  stress,  loS. 
Bevel-wheels,  335. 
Boilers,  586. 
Bollman  truss,  56,  618. 
Bowstring  truss,  61,  618. 
Brace,  i,  25. 
Brakes,  323. 

Breaking-down  point,  149. 
Breaking  stress,  147. 

weights,  343,  399. 
Breaking  weights  of  iron  girders,  369, 

370. 
Breaking  weiKhis,  tables  of,  212. 
Brickwork,  149. 

Bridge,  bowstring  suspension,  626. 
"        loads,  600. 
"        trusses,  17,  52. 
"      '      ''        chords  of,  625. 
"  "         depth  of,  597. 

Bridge    trusses,     maximum    allowable 

stress  in,  657. 
Bridge  trusses,  stiffness  of,  598. 
stringers  of,  656. 
Bridges,  597. 

"         position  of  platform  of,  598. 
Buckling  of  pillars,  513,  515. 

813 


8X4 


INDEX. 


Cable  with  sloping  suspenders,  717. 
Cables,  703. 

"       curves  of,  706. 
"      deflection  of,  714. 
"      length  of  arc  of,  712. 
"       parameter  of,  711, 
"       weight  of,  713. 
Camber,  3SS,  659.  v  • 

Cantilever,  365. 

"  curve  of  boom  of,  634. 

"  deflection  of,  638. 

"  depth  of   637. 

"  weight  of,  632. 

Cast-iron,  147. 
Catenary,  34,  706,  750. 
Cement,  150. 
Centres  of  gravity,  11. 
Centre  of  resistance,  I,  743. 
Centrifugal  force,  iSi. 
Centripetal  force,  1S2. 
Clapeyron's  theorem,  292. 
Coefficient  of  cu>  ic  elasticity,  255. 
"  "  e'  isticity,   141,  143. 

"  "  f  uidity,  162. 

"  "    lardness,  164. 

"  "  i^'eral  elasticity,  144. 

"  "  rigidity,  254,  285. 

"  "  rupture,  248. 

"  "  torsional  rupture,  574. 

"  "  transverse  elasticity,  285. 

Collar-beams,  25. 
Columns,  see  Pillars,  513,  53S. 
flexure  of,  554,  557. 
Compound  strain,  236. 
Compression,  141. 
Conjugate  stresses,  247. 
Continuous  girders,  463. 
Continuous  girders,  advantages  and  dis- 
advantages of,  486. 
Continuous  girders,  maximum  bending 

moment  in,  465. 
Coulomb's  laws,  568. 
Counterbrace,  60. 
Counter-efficiency,  328, 
Counterforts,  270. 
Covers  of  riveted  joints,  665. 
Cranes,  13. 

"        bent,  31. 

"        derrick,  16.  ■      ' 

"       jib,  13. 
pit,  15. 
Crank  effort,  207. 
Cubic  elasticity,  255. 
"      strain,  283. 

Dead  load,  143,  600. 
Deflection,  curve  of,  434. 

of  girde..,  384-386,  638. 
Deformation,  140,  251,  254. 


Dock  walls,  270. 
Dynamometer,  Prony's,  327. 

Earth  foundations,  258. 
Earthwork,  255. 

''  pressure  of,  257. 

Earthwork,  Rankine's  theory  applied  to 

retaining  walls,  264. 
Efficiency  of  mechanisms,  335. 

"         of  riveted  joints,  606. 
Elastic  curve,  355. 

"       moment,  96,  340. 
Elasticity,  140. 

coefficient  of,  '.41,  143. 
'•  "  "  cubic,  255. 

'■  "  "  lateral,  144. 

"  1.         .<  transverse,  285. 

"  limit  of,  145. 

Ellipse  of  stress,  241. 
Ellipsoid  of  stress,  281. 
Empirical  rules  for  wind-pressure,  663. 
Encasire  girders,  458. 
Energy,  207. 

"        curves  of,  207. 
"        fluctuation  of,  207. 
"        kinetic,  167,  169,  170. 
"        potential,  167. 
Envelope  of  moments,  121. 
Equalization  of  stress,  349. 
Equalizer,  629. 
Equilibrated  polygon,  740, 
Equilibrium  of  bea.ms,  428. 
Equilibrium  of  beams,  general  equations 

of,  428. 
Equilibrium  of  flanged  girders,  366. 
Euler's  theory  of  the  strength  of  pillars, 

537- 
Examples,     69-92,      132-139,     21(1-234, 

294-205,  337-339.  407-427.  4QO-5I2, 
'56?-567.  5S>o-5S5,  594-59^.  (>89-702. 
734-739.  S09-812. 

Expansion  of  solids,  215. 

Extension  of  prismatic  bar,  289. 

Extrados,  740. 

Eyebars,  661. 

"  steel,  665. 

Factor  of  safety,  150. 
Fatigue,  152. 
Fink  truss,  54. 
Flanged  girders,  365. 

"  "        equilibrium  of,  36b. 

"  "        stiffness  of,  384. 

Flanges,  365,  597. 

"        curved,  366. 

"        horizontal.  366. 
Flexure    of    columns,  see  Pillars,  554, 

557. 
Flow  of  solids,  162. 


INDEX. 


815 


Fluctuation  of  stress,  151. 
Fluid  pressure,  162. 
Force  polygon,  3,  7,  119. 
Foundations,  earth,  258. 

"  limiting  depth  of,  258. 

"  of  walls,  270. 

Fracture,  141. 

Framed  arch,  stresses  in,  804. 
Framed  arch.  Clerk  Maxwell's  method 

of  determining  stresses  in  a,  806. 
Frames,  i,  2. 

"        incomplete,  27,  61. 
Friction,  300. 

angle  of,  237. 
"         coefficient  of,  300,  313. 
"         journal,  310. 
"  rolling,  310. 

Funicular  curve,  10. 

polygon,  3,  7,  117,  119. 

Gins,  17. 

Girder  of  uniform  strength,  381. 

Gordon's  formulfe  for  pillars,  522. 

Hinged  girders,  127-131. 
Hodgkinson's  formulae  for  the  strength 

of  pillars,  513,  517-521. 
Hooke's  law,  142. 
Howe  truss,  58,  6ir. 

Impact,  184. 
Impulse,  176. 
Incomplete  frames,  27. 
Inertia,  198. 

"       moment  of,  12,  342, 

"        pressure  due  to,  200. 
Inflection,  point  of,  453-463. 
Internal  stress,  235. 
Isotropic  bodies,  283. 

Joint  of  rupture,  747, 

Keystone,  741. 

Lateral  bracing,  654. 
Lattice  girder,  600. 
Launhardt's  formula,  153. 
Lenticular  truss,  626. 
Limit  of  elasticity,  145. 
Line  of  loads.  5. 

"     "  resistance,  273-276,  741,  750. 

"     "  rupture,  265. 
Linear  arch,  743,  753-760. 
Loads,  live,  iii,  115,  119,  600,639,641, 

730. 
Loads,  stationary  (dead),  118,  600, 
Long  pillars,  535. 

Mansard  roof,  6. 

Masonry,  149. 

Mechanical  advantage,  294. 


Middle  third  theory  of  arches,  746. 
Modulus  of  (ilasticity,  141. 

"        "  rupture,  348. 

"        "  transverse  elasticity,  254. 
Moment  of  forces,  n6 

"        "  inertia,  12,  342. 

"       "      "       examples  of,  371-81. 
Moment  of  inertia,  variable  section   of, 

455. 
Moment  of  inflexibility,  96. 

"        "  resistance,  96, 
Momentum,  176. 
Mortar,  150. 

Neutral  axis,  340. 

"     of  a  loaded  beam,  435-454. 
"  "     surface,  340. 

Oblique  resistance,  169. 
Oscillatory  motion,  190,  195. 

Panel  points,  52. 
Panels,  54. 
Piers,  65. 
Pillars,  513. 

"        Euler's  formulae  for,  527. 

"         failure  of,  515. 

"         flexure  of,  515. 

"         formulae  for  American,  527-532. 

"        Gordon's  formulae  for,  522. 
Pillars,  Hodgkinson's  formulae  for,  517- 

521. 
Pillars,  Rankine's  formula  for,  526. 
Pillars  with  stress  uniformly  distributed. 

516. 
Pillars  with  uniformly   varying   stress, 

517- 
Pins,  661. 

Piston  velocity,  curves  of,  205. 
Pivots,  316. 

"       conical,  319 

"       cylindrica. ,  316. 

Schieie's  (anti-friction),  320. 
Plasticity,  141. 

Poisson's  ratio,  142. 

Pole,  7. 

Polygon  of  forces,  3,  7. 

"         "  pressure,  743. 
Pratt  truss,  60. 
Primitive  strength,  153, 
Principals,  33,  34. 
Prony's  dynamometer,  327. 
Proof  strain,   171. 

"     stress,  171. 
Purchase   304. 
Purlins,  33,  34. 

Radius  of  gyration,  174,  528-531. 
Rale  of  twist,  289. 


8i6 


INDEX. 


Redundant  bars,  48. 

Reservoir  walls,  271. 

Resilience,  171, 

Retaining  walls,  260. 

Retaining  walls,   conditions   of   equiib- 

rium  of,  260. 
Retaining  walls,    Rankine's   theory   ap- 
plied 10,  264. 
Rivet   connection    bef.veen    flange   and 

web,  660. 
Riveted  joints,  668 

"  "       covers  of,  675. 

•«  "      design  of,  67S. 

"  "       efficiency  oi',  679. 

"  "       failure  of,  670. 

"  "      stresses  in,  670. 

"  "       theory  of,  671-675. 

Riveting,  666. 
RivetE,  666. 

"       dimensions  of,  667. 
Rocker-link,  629. 
Rollers,  35,  639. 
Roof  trusses,  17. 

"        "        distributionof  loadson,  39. 
"         types  of,  33. 

'  "         weights  of,  37. 

Ropes,  321. 


Saddles,  704. 
S-^hiele's  pivots,  320. 
Screws,  306. 

"        endless,  309. 
Sections,  method  of,  62 
Set,  145. 
Shafting,  distance  between  bearings  of, 

575- 
Shafting,  efficiency  of,  577. 

"         internal  stress  in,  237. 

"         stiffness  of,  573. 

"         torsional  strength  of,  288-291. 
Shear,  141. 

"      maximum,  121,  237. 
Shearing  force,  95. 

"  "      examples  of,  97,  108. 

Shearing  force   and  bending    moment, 

relation  between,  108. 
Shearing  stress,  19S. 

"  "        distribution  of,  391. 

Shear-legs,  17. 
Similar  girders,  401-404. 
Skew-backs,  34,  740. 
Soffit,  740. 
Spandril,  740. 
Specific  weight,  143. 
Spherical  shells,  591. 
Spritigings,  740. 
Springs,  355,  45M5S. 

"        simple  rectangular,  456. 


Springs,  spiral,  477. 

Springs  of  constant  depth,  but  triangu- 
lar in  plan,  457. 
Springs  of  constant  width,  but  parabolic 

in  elevation,  457. 
Statical  breaking  strength,  153. 
Steel,  148. 
Stiffening  truss,  719, 

"      hinged  at  centre,  725. 
Stiffness,  190,  387,  389. 
Strain,  140,  251. 
Straining  cill,  18. 
Stress,  141,  251 

"      and  strain,  relation  betw«en,  281. 

"       general  equations  of,  277. 

"       principal,   241. 

"  "  planes  of,  237. 

Stresses,  conjugate,  248,  250. 
Stress-strain  curves,  147-149. 

"  line,  144. 

Strut,  r. 

St.  Venant's  torsion  results,  572. 
Surface  loading,  350. 
Suspenders,  706. 
Suspension-bridges,  703. 

"  loads  on,  730. 

Suspension-bridges,    modifications     of, 

731- 
Suspension-bridges,  pressure  upon  piers 

of,  718. 
Swing-bridges,  470-472. 

Tables  of  breaking  weights  and  coeffi- 
cients of  bending  strength  of  limber, 
212,  213. 

Table  of  coefficients  of  axle  friction, 
336. 

Table  of  coefficients  in  Gordon's  for- 
mula, 524. 

Table  of  coefficients  in  Rankine's  mod- 
ification of  Gordon's  formula,  526. 

Tables  of  diagonal  and  chord  stresses, 
644-650. 

Tables  of  efficiencies,  587. 

"      "  elliptic  integrals,  562. 
"      "  expansion  of  solids,  215. 
"      "  eyebar  dimensions,  665. 
"      "  factors  of  safety,  214. 

Tables  of  loads  for  highway  bridges, 
687. 

Tables  of  strengths,  elasticities,  and 
weights  of  iron  and  steel,  210. 

Table  of  strengths,  elasticities,  and 
weights  of  various  alloys,  211, 

Tables  of  weights  of  modern  bridges, 
682-687. 

Table  of  weights  and  crushing  strength 
of  rocks,  214 

Table  of  weights  of  roof  coverings,  67. 


INDEX. 


817 


Table  of  weights  of  roof  frames,  67. 

Tension,  141. 

Theorem  of    hree  moments,  463-470. 

Thick  hollou  cylinder,  588. 

Timber,  149, 

Torsion,  141,  568. 

St.  Venani's  results,  572. 
Torsional  coefficient  of  elasticity,  145. 

"  resilience,  574. 

Torsional  strength  of  shafts,  2S8,  280, 

571,  572. 
Transverse  strength,  141. 
Transverse  vibration  of  a  loaded  beam 
461.  ' 

Trellis  girder,  600. 

Tresca's  theory  of  flow  of  solids,  162. 
Tripods,  17. 
Truss,  2. 

"      composite,  31. 
king-post,  21. 

"      queen-post,  25. 

"      roof,  32. 

"      triangular,  19, 
Trussed  beams,  53, 
Twist,  141. 

Unwin's  formula,  159. 

Values  of  /6*.  174,  528-531. 
Vibration  strength,  153. 
Voussoir,  741, 

Warren  truss,  57. 


Web  thickness,  382. 

Wedge,  303 

Weights  of  roof  coverings,  67. 

"       "    frames,  67. 
Weyrauch's  formula,  153. 
Weyrauch's  theory  of  buckling  of  pillars. 

550. 
Wheel  and  axle,  329. 
Whipple  truss,  618. 
Wind  pressure,  38,  67,  629,  651,  653. 
Wmd-pressure,  American  specifications 

of,  652. 
Wind-Pressure  Commission  rules,  653. 
Wind-pressure,   empirical    regulations, 

653- 
Wohler's  law,  150. 
Work,   167. 

"       effective,  178. 
"       external,  168. 
"       internal,  168. 
"       useful,  178. 
"       waste,  178. 
Work  done  in  bending  a  beam,  460. 
Work  done  in    small   deformation  of  a 

body,  292. 
Work  of  journal  friction,  314. 
Working  load,  150. 

strength,  150. 
"         stress,  150, 
Wrought-iron,  148. 

Yield-point,  149. 


